Report 5
Intermediate Engineering Analysis
Section 7566
Team 11
Due date: March 30, 2012.
Given: Find
for the following series:
1. ![{\displaystyle r(x)=\sum _{k=0}^{\infty }(k+1)kx^{k}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ac337e7de07b7984f2f1b6ba956b64c009d6ceb)
2. ![{\displaystyle r(x)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{\gamma ^{k}}}x^{2k}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a6523e0764c9bd81a5c2febfed4e60398727ebd)
Find
for the Taylor series of
3.
at ![{\displaystyle x=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4921ebe9c3cbb5bf9412d05da9f40c4ac085801)
4.
at ![{\displaystyle x=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4921ebe9c3cbb5bf9412d05da9f40c4ac085801)
5.
at ![{\displaystyle x=1\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d5649865ee476680d4627e431dada6217962d99)
The radius of convergence
is defined as
![{\displaystyle R_{c}=[\lim _{k\to \infty }|{\frac {d_{k+1}}{d_{k}}}|]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6b8ccc715a7471923f69e28178706700849ca1f)
1.
![{\displaystyle R_{c}=[\lim _{k\to \infty }|{\frac {(k+2)(k+1)}{(k+1)(k)}}|]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc71081c7afee94390e37f71f56d2193dec48de9)
![{\displaystyle R_{c}=[\lim _{k\to \infty }|{\frac {(k+2)}{(k)}}|]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1af688ca1f47d693121b1271218fd929490e415)
![{\displaystyle R_{c}=[\lim _{k\to \infty }|{\frac {k(1+{\frac {2}{k}})}{k}}|]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91507d0db845548c113c47343de793dc125cf834)
![{\displaystyle R_{c}=[\lim _{k\to \infty }|1+{\frac {2}{k}}|]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c8a3989bc7fc0b444be04b556b2d2ee1c1a1d7d)
![{\displaystyle R_{c}=[1+0]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea6fe870b92b59fdd2dd61e493f66c583d007ff5)
![{\displaystyle R_{c}=1\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9630c60c3f3b32d25e7dcae16c8c6b8c7b372cd7)
2.
![{\displaystyle R_{c}=[\lim _{k\to \infty }|{\frac {\frac {(-1)^{k+1}}{\gamma ^{k+1}}}{\frac {(-1)^{k}}{\gamma ^{k}}}}|]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05386557debdb49583acce7ab9f83a1d4be70b88)
![{\displaystyle R_{c}=[\lim _{k\to \infty }|{\frac {-1}{\gamma }}|]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4544068f0608761e16857dd77cca7662123056f)
![{\displaystyle R_{c}=[\lim _{k\to \infty }{\frac {1}{\gamma }}]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7997241d23909c23c91497322fbe1ff149d57d78)
![{\displaystyle R_{c}=[{\frac {1}{\gamma }}]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1414ef0ffee2c582ed3eeaf9a7993a628b08b4d0)
![{\displaystyle R_{c}=\gamma \!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fa399b7c50b94ee336f583a16dddcfee58c7dd3)
However, in this problem, the series
term is
not
, as is the general form.
Therefore, this implies:
![{\displaystyle |x^{2}|=\gamma \!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9789d2586382370585d57921e19fba5266a3a5c9)
![{\displaystyle |x|={\sqrt {\gamma }}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca82e2b4ca7aa76fef28417a9593e19ad0aef102)
![{\displaystyle R_{c}={\sqrt {\gamma }}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f24cd5464cdd18192e69623fe4eaabc3a557d12)
3. The Taylor series for
is expressed as ![{\displaystyle sin(x)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}x^{2k+1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c81ac953dd6ca01b5d1c01548d89c29e5770f78b)
![{\displaystyle \lim _{k\to \infty }{\frac {{\frac {(-1)^{k+1}}{(2(k+1)+1)!}}x^{2(k+1)+1}}{{\frac {(-1)^{k}}{(2k+1)!}}x^{2k+1}}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c427f63f735013c3c95b7b84aee86442bfbae4f0)
![{\displaystyle \lim _{k\to \infty }{\frac {{\frac {1}{(2k+3)!}}x^{2k+3}}{{\frac {1}{(2k+1)!}}x^{2k+1}}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d318503befeaaaaa5c00a8eb73a172c32288793b)
![{\displaystyle \lim _{k\to \infty }{\frac {{\frac {1}{(2k+3)}}x^{3}}{x}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f6b11f84f12111458cd7afb95656e3c9c93da28)
![{\displaystyle \lim _{k\to \infty }{\frac {1}{(2k+3)}}x^{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ec8073aa9396291af1b8dc19d25adf07503ade4)
Therefore:
![{\displaystyle R_{c}=[\lim _{k\to \infty }{\frac {1}{(2k+3)}}]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b3f749928e2e6110c615a89db49b3e83b1ca1c5)
![{\displaystyle R_{c}=\infty \!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9cdaa5156b656b7413917a2ddd209ec788441839)
4. The Taylor series for
at
is expressed as ![{\displaystyle log(1+x)=\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}x^{k}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e1b40322b09ea6fbd9b17101638ad2384466a2b)
![{\displaystyle R_{c}=[\lim _{k\to \infty }|{\frac {\frac {(-1)^{k+2}}{k+1}}{\frac {(-1)^{k+1}}{k}}}|]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c89f760e5fa1462fac57cfd53e47c7ba4695f4d)
![{\displaystyle R_{c}=[\lim _{k\to \infty }|{\frac {\frac {(-1)}{k+1}}{\frac {(1)}{k}}}|]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0fd04d810dc90cb7de776420e8b664f3ff32f6f)
![{\displaystyle R_{c}=[\lim _{k\to \infty }|{\frac {\frac {(-1)}{k(1+{\frac {1}{k}})}}{\frac {1}{k}}}|]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/27d01121f5bf24b5b3c0f00bdc9fe4cea9fabdaf)
![{\displaystyle R_{c}=[\lim _{k\to \infty }|{\frac {(-1)}{(1+{\frac {1}{k}})}}|]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de135421582f0cd5c0506882a31fb79c9d191882)
![{\displaystyle R_{c}=[1/1]^{-1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18d3becf975c7c5cb357e14fbb8da3130baba3f1)
5. The Taylor series for
at
is expressed as ![{\displaystyle log(1+x)=log(2)+\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{2^{k}k}}(x-1)^{k}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de0acd420cd2dd056e54979ca97fb6270886f59f)
![{\displaystyle \lim _{k\to \infty }{\frac {{\frac {(-1)^{k+2}}{2^{k+1}(k+1)}}(x-1)^{k+1}}{{\frac {(-1)^{k+1}}{2^{k}k}}(x-1)^{k}}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c2339dad46988d9bdd8a633066f3f7fe178aff0)
![{\displaystyle \lim _{k\to \infty }{\frac {{\frac {(-1)^{2}}{2(k+1)}}(x-1)}{\frac {1}{k}}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6587ff9d05db6e062f4ded17fecf4c5ca75fa1e)
![{\displaystyle \lim _{k\to \infty }{\frac {{\frac {1}{2k(1+{\frac {1}{k}})}}(x-1)}{\frac {1}{k}}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/121e20932c51fefcca9cb40e8991f922f5728455)
![{\displaystyle \lim _{k\to \infty }{\frac {{\frac {1}{2(1+{\frac {1}{k}})}}(x-1)}{1}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5527ea1be5f65127992f644d99bcf400ec01dc8a)
![{\displaystyle {\frac {1}{2}}(x-1)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/437ccdc68bda77be651de265d72d52b1c534d688)
For convergence: ![{\displaystyle {\frac {1}{2}}(x-1)<1\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e2b3ae72dd864c10ce4274f67ad11f024a780f8)
![{\displaystyle x-1<2\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/17799c7f4529164420a8a62837bd62dfc8e8803e)
![{\displaystyle x<3\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53e10d46ecc192adff98f5dbf55e1a04eac3b365)
Therefore,
Solved by: Andrea Vargas
Part 1:Determine whether the following are linearly independent using the Wronskian
Part 2: Determine whether the following are linearly independent using the Gramian
Using the Wronskian we check for linear independence.
We know from (1) and (2) in 7-35 that if
![{\displaystyle W=\det {\begin{bmatrix}f&g\\f'&g'\end{bmatrix}}=fg'-gf'\neq 0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4231363c7b5e2926360d236d49d129014e36ca5)
Then the functions are linearly independent.
![{\displaystyle f(x)=x^{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8abf3cde454dbfcfb3e35545248c0c859fa3c8d)
![{\displaystyle g(x)=x^{4}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06c31d126d453f0f7783d76332c2c181cb80a178)
Taking the derivatives of each function:
![{\displaystyle f'(x)=2x\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/912717062dc288b91dc56cd49753be26b6e75234)
![{\displaystyle g'(x)=4x^{3}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b175f089d9cc1233508e51ceffe8cd79d6078fd)
![{\displaystyle W={\begin{bmatrix}x^{2}&x^{4}\\2x&4x^{3}\end{bmatrix}}=4x^{5}-2x^{5}=2x^{5}\neq 0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a5125c5ccb13b3e22ef7779adeafdb9119e2ac4)
f(x) and g(x) are linearly independent
![{\displaystyle f(x)=\cos(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a75209a1524717f8a745dd0f65751b4ebdc8358)
![{\displaystyle g(x)=\sin(3x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/73c6f9ef92cc1c7a043ee9270d2fcc70f2eeb1bd)
Taking the derivatives of each function:
![{\displaystyle f'(x)=-\sin(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20e81ce8e5e797e0d2caa03942fe533d920c0736)
![{\displaystyle g'(x)=3\cos(3x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d9d38785f5c849ee2b7315e63a1605dac361355)
![{\displaystyle W={\begin{bmatrix}\cos(x)&\sin(3x)\\-\sin(x)&3\cos(3x)\end{bmatrix}}=3\cos(x)\cos(3x)+\sin(3x)\sin(x)=\neq 0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c93d66cb9096f6d9190c8875a49d449661f57f4)
f(x) and g(x) are linearly independent
Using the Gramian we check for linear independence.
We know from the notes in (1) 7-34 that:
![{\displaystyle \langle f,g\rangle :=\int _{a}^{b}=f(x)g(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff4b35ec8b7f2721f7e980ffc4f5520a41e2acfe)
and that the Gramian is defined as:
Then f,g are linearly independent if ![{\displaystyle \Gamma (f,g)\neq 0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8930ca9fc3a69a4a73b99276eb09dcea032cff7)
![{\displaystyle f(x)=x^{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8abf3cde454dbfcfb3e35545248c0c859fa3c8d)
![{\displaystyle g(x)=x^{4}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06c31d126d453f0f7783d76332c2c181cb80a178)
Taking scalar products:
![{\displaystyle \langle f,f\rangle :=\int _{-1}^{1}=f(x)f(x)=x^{2}*x^{2}=\int _{-1}^{1}=x^{4}=\left[{\frac {x^{5}}{5}}\right]_{-1}^{1}={\frac {2}{5}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9558e9969a77a0045d0716d1f5ae91e7ddad89e)
![{\displaystyle \langle f,g\rangle =\langle g,f\rangle :=\int _{-1}^{1}=f(x)g(x)=x^{2}*x^{4}=\int _{-1}^{1}=x^{6}=\left[{\frac {x^{7}}{7}}\right]_{-1}^{1}={\frac {2}{7}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/602b892d340a3d3b15cdad195048562b0b4c3d81)
![{\displaystyle \langle g,g\rangle :=\int _{-1}^{1}=g(x)g(x)=x^{4}*x^{4}=\int _{-1}^{1}=x^{8}=\left[{\frac {x^{9}}{9}}\right]_{-1}^{1}={\frac {2}{9}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7ea937455865cedfe397a7909635c90e2c5c920)
![{\displaystyle \Gamma (f,g)=\det {\begin{bmatrix}{\frac {2}{5}}&{\frac {2}{7}}\\&\\{\frac {2}{7}}&{\frac {2}{9}}\end{bmatrix}}=0.08888-0.0816326531\neq 0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7314dff2a8d64ef5e673bca30151ecb6e2636aaf)
f(x) and g(x) are linearly independent
![{\displaystyle f(x)=\cos(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a75209a1524717f8a745dd0f65751b4ebdc8358)
![{\displaystyle g(x)=\sin(3x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/73c6f9ef92cc1c7a043ee9270d2fcc70f2eeb1bd)
Taking scalar products:
![{\displaystyle \langle f,f\rangle :=\int _{-1}^{1}=f(x)f(x)=\int _{-1}^{1}=\cos ^{2}(x)dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3b6471f7519838223d04906b301e52f6e95a262)
We can use the trig identity for power reduction ![{\displaystyle \cos ^{2}(x)={\frac {1}{2}}\cos(2x)+{\frac {1}{2}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/58dc0c853b31ca1c25044c72af8833802be5b14b)
Then we have,
![{\displaystyle \int _{-1}^{1}{\frac {1}{2}}\cos(2x)+{\frac {1}{2}}dx=\left[{\frac {1}{4}}\sin(2x)+{\frac {1}{2}}x\right]_{-1}^{1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b430bf3d0b57d4d80031e1c87c89e7e84cf1a67)
![{\displaystyle =0.7273243567-(-0.7273243567)=1.454648713\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71b3a1759eb4baaa6cd825a1d97969ca7a3a376c)
![{\displaystyle \langle f,g\rangle =\langle g,f\rangle :=\int _{-1}^{1}f(x)g(x)dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d98d5af04b3e1171636b89b727ef22a11c68351)
![{\displaystyle =\int _{-1}^{1}\sin(3x)\cos(x)dx=\left[{\frac {1}{8}}(-2\cos(2x)-\cos(4x))\right]_{-1}^{1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8203ee3878256536c05c42f6634ab0d963539e6e)
![{\displaystyle =0.185742-0.185742=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee59d920d2e65bb2041499da13128647c06b978e)
![{\displaystyle \langle g,g\rangle :=\int _{-1}^{1}g(x)g(x)dx=\int _{-1}^{1}\sin(3x)\sin(3x)dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6ae0dcbb92cf6584fedb14ce741311cc358597f)
![{\displaystyle ={\frac {1}{2}}\int _{-1}^{1}1-\cos(6x)dx=\left[{\frac {1}{2}}(x-{\frac {1}{6}}\sin(6x))\right]_{-1}^{1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ce4d0a3d02f4039abbee99ac372ff1fb6572374)
![{\displaystyle =0.523285+0.523285=1.04656925\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/401d39f35e309cf9bace640de52820ab33b613f4)
![{\displaystyle \Gamma (f,g)=\det {\begin{bmatrix}1.454648713&0\\0&1.04656925\end{bmatrix}}=1.522390612\neq 0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1200bda18b711791315730137675cf8fc85c58de)
f(x) and g(x) are linearly independent
By both methods (the Wronskian and the Gramian) we obtain the same results.
Verify using the Gramian that the following two vectors are linearly independent.
![{\displaystyle \mathbf {b_{1}} =2\mathbf {e_{1}} +7\mathbf {e_{2}} \!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4594b6fb758357deb2c7d4c62b91a0d0fa7df639)
![{\displaystyle \Gamma (f,g)=det{\begin{bmatrix}\langle \mathbf {b1} ,\mathbf {b1} \rangle &\langle \mathbf {b1} ,\mathbf {b2} \rangle \\\langle \mathbf {b2} ,\mathbf {b1} \rangle &\langle \mathbf {b2} ,\mathbf {b2} \rangle \end{bmatrix}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6acb81074169c03e4d58c33eb179e3d2ee0fc71)
We know from (3) 8-9 that:
We obtain,
![{\displaystyle \langle \mathbf {b1} ,\mathbf {b1} \rangle =(2)(2)+(7)(7)=4+49=53\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d58b6ab4f9514f8007b725f04e1163d132d7df8)
![{\displaystyle \langle \mathbf {b2} ,\mathbf {b2} \rangle =(1.5)(1.5)+(3)(3)=2.25+9=11.25\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d66b256b521c2bf3b8d3d26ae08f59e4660597f0)
Then,
![{\displaystyle \Gamma (f,g)=det{\begin{bmatrix}53&24\\24&11.25\end{bmatrix}}=596.25-576=20.25\neq 0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f17e4e97c0f2fbcb78b1893499d1ef92ce365013)
b_1 and b_2 are linearly independent
Show that
is indeed the overall particular solution of the L2-ODE-VC
with the excitation
.
Discuss the choice of
in the above table e.g., for
why would you need to have both
and
in
?
Because the ODE is a linear equation in y and its derivatives with respect to x, the superposition principle can be applied:
is a specific excitation with known form of
and
is a specific excitation with known form of
becomes
proving that
is indeed the overall particular solution of the L2-ODE-VC
with the excitation
According to Fourier Theorem periodic functions can be represented as infinite series in terms of cosines and sines:
where the coefficients
are the Fourier coefficients calculated using Euler formulas.
So even though the system is being excited by functions like
the particular solution would still include both
and
in
because the excitation is a periodic function that can be represented as the Fourier infinite series in terms of both
and
times the Fourier coefficients
Show that
and
are linearly independant using the Wronskian and the Gramain (integrate over 1 period)
![{\displaystyle f=cos(7x),g=sin(7x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e6d7beab3a98af7b5cba6940e421659fc28f8a0e)
One period of ![{\displaystyle 7x=\pi /7}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4278f5c13873c154cd6634f28e43ded580f9e5e1)
Wronskian of f and g
![{\displaystyle W(f,g)=det{\begin{bmatrix}f&g\\f'&g'\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47dfe8cfd362e670f7fcc60915f0ecd5235addbe)
Plugging in values for ![{\displaystyle f,f',g,g';}](https://wikimedia.org/api/rest_v1/media/math/render/svg/46abef6dc84fa8fe963aefdff17a5bf1ba40a1fc)
![{\displaystyle =7cos^{2}(7x)+7sin^{2}(7x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f8b0e124ec60f773018903a5ffb0eb1318cb2cf)
![{\displaystyle =7[cos^{2}(7x)+sin^{2}(7x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b301de567d6ca791f8580ee8e9e260a31d103757)
![{\displaystyle =7[1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b92b8a5eea92d540f3a060ab9d154589aa005e48)
They are linearly Independant using the Wronskian.
![{\displaystyle <f,g>=\int _{a}^{b}f(x)g(x)dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/936d91732c7f2d47bfe4dad4fcca8007e432b661)
![{\displaystyle \Gamma (f,g)=det{\begin{bmatrix}<f,f>&<f,g>\\<g,f>&<g,g>\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91f89b547a18369fe7bf2d7f558a02f5241f26e1)
![{\displaystyle \int _{0}^{\pi /7}cos^{2}(7x)dx=\pi /14}](https://wikimedia.org/api/rest_v1/media/math/render/svg/efba1a93c8bb7b43198d57b0868e58bcf08787f5)
![{\displaystyle \int _{0}^{\pi /7}sin^{2}(7x)dx=\pi /14}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c263bc98e45e25490399794f9aaa94ab3be2956)
![{\displaystyle \int _{0}^{\pi /7}cos(7x)*sin(7x)dx=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44c37749ac7121f5c2e0ae44104e50395189c68e)
![{\displaystyle \Gamma (f,g)=det{\begin{bmatrix}\pi /14&0\\0&\pi /14\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42a7ef16d7aee1475459f1d05dd987c2ef2b73c8)
They are linearly Independent using the Gramain.
Find 2 equations for the 2 unknowns M,N and solve for M,N.
![{\displaystyle y_{p}(x)=Mcos7x+Nsin7x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fbd1902ad6b6aba1cd6cba7a2ee31e91de27acac)
![{\displaystyle y'_{p}(x)=-M7sin7x+N7cos7x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e042d90da812df623a011eecc813d91e03a29c87)
![{\displaystyle y''_{p}(x)=-M7^{2}cos7x-N7^{2}sin7x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdc780d6943b5347afc4a0bdc6710b8b4628969e)
Plugging these values into the equation given (
) yields;
![{\displaystyle -M7^{2}cos7x-N7^{2}sin7x-3(-M7sin7x+N7cos7x)-10(Mcos7x+Nsin7x)=3cos7x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/82c56c74161c2d47b0a5b211ec45aa5f9461526d)
Simplifying and the equating the coefficients relating sin and cos results in;
![{\displaystyle -59M-21N=3}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a545e0454310c32ee32fe8ebe9ace0f56a8ec25)
![{\displaystyle -59N+21M=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41599723c770b384172beb583ffa875e7933069b)
Solving for M and N results in;
![{\displaystyle M=-177/3922,N=-63/3922}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b165ca4138c70bc7c60c68efce4cc51e06ad49b)
Find the overall solution
that corresponds to the initial conditions
. Plot over three periods.
From before, one period
so therefore, three periods is ![{\displaystyle 3\pi /7.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bbee26e8d1cd06196952022decf05b8c3ad165c7)
Using the roots given in the notes
, the homogenous solution becomes;
![{\displaystyle y_{h}(x)=c_{1}e^{-2x}+c_{2}e^{5x}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6722fad8e628c9602e377767e8eabae9ab9056c0)
Using initial condtion
;
![{\displaystyle 1=c_{1}+c_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ddb8e2e309bd352775037bba1b4063e7ce3d5fd3)
with ![{\displaystyle y'(0)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13d7851b0db9f542863ddbbc0f4875874e6bac12)
![{\displaystyle 0=-2c_{1}+5c_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5cae60cc88b2e6d1852e7d5b3c46f64b0534f795)
Solving for the constants;
![{\displaystyle c_{1}=5/7,c_{2}=2/7}](https://wikimedia.org/api/rest_v1/media/math/render/svg/231969602e15abd8d16d69ff1b4dcc8228c4ceef)
![{\displaystyle y_{h}(x)=5/7e^{-2x}+2/7e^{5x}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05236e08c168c45ea4d34d76d39ccd3abf8476ee)
Using the
found in the last part;
![{\displaystyle y=y_{h}+y_{p}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2756dd440f8cfa429a60bf180c123bdf6bf949c6)
![](//upload.wikimedia.org/wikiversity/en/f/fd/R5_code.jpg)
solved by Luca Imponenti
Complete the solution to the following problem
![{\displaystyle y''+4y'+13y=2e^{-2x}cos(3x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa8856fe667dfd446436d60b60e40b2c4fd97c63)
where
and
Find the overall solution
corresponds to the initial condition:
![{\displaystyle y(0)=1\ ,\ y'(0)=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/583283c322ca54967e07641dbc6f6d3715cebe73)
Plot the solution over 3 periods.
Taking the derivatives of the particular solution
Plugging these into the ODE yields
Equating like terms allows us to solve for M and N
So the particular solution is
The overall solution in the sum of the homogeneous and particular solutions
To find A and B we apply the initial conditions
Taking the derivative
Giving us the overall solution
The period for
is
Plotting the solution
over 3 periods yields
Solved by Daniel Suh
![{\displaystyle v=4e_{1}+2e_{2}=c_{1}b_{1}+c{2}b_{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ca6363c1ffb79332d3ad352f9456b62608b332b)
![{\displaystyle b_{1}=2e_{1}+7e_{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e99649647da4f84900a60f39c42c9dec6aa5c31)
![{\displaystyle b_{2}=1.5e_{1}+3e_{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/587c2a7b73c89b8de53f3fd03b30c6165ad1dcc5)
1. Find the components
using the Gram matrix.
2. Verify the result by using
and
, and rely on the non-zero determinant matrix of
and
relative to the bases of
and
.
![{\displaystyle T(b_{1},b_{2})={\begin{bmatrix}<b_{1},b_{1}>&<b_{1},b_{2}>\\<b_{2},b_{1}>&<b_{2},b_{2}>\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af9422363295abdd7b2f65ce112669a590eca57a)
![{\displaystyle <b_{i},b_{j}>=<b_{i}\cdot b_{j}>\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0fb4d1565736b0e9d7ae6faac3dcca79b031866e)
Thus,
![{\displaystyle <b_{1},b_{1}>=<b_{1}\cdot b_{1}>=<(2)(2)+(7)(7)>=53\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69b037acb4ec2293d019c6f7b9c65263c250e757)
![{\displaystyle <b_{2},b_{2}>=<b_{2}\cdot b_{2}>=<(1.5)(1.5)+(3)(3)>=11.25\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f28dd36883930c9af71992814fb4c79f3ff6da8)
![{\displaystyle <b_{1},b_{2}>=<b_{2},b_{1}>=<b_{1}\cdot b_{2}>=<(2)(1.5)+(7)(3)>=24\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d100eff479c9fe10ea28019621e681a0f0aa8b3)
![{\displaystyle T(b_{1},b_{2})={\begin{bmatrix}53&24\\24&11.25\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f3df7bd3ae7e1dd6cde03a4db74f391d87f2b537)
![{\displaystyle \Gamma =det[T]=(53)(11.25)-(24)(24)=20.25\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a18e1ee1c3750d81d34b124c3b39230952ce917)
Define:
![{\displaystyle c={\begin{bmatrix}c_{1}\\c_{2}\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ffd17d6f30a74b580cf63ccd0b34b3d20e901ae)
![{\displaystyle d={\begin{bmatrix}<b_{1},v>\\<b_{2},v>\end{bmatrix}}={\begin{bmatrix}d_{1}\\d_{2}\end{bmatrix}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/58ad4ade098195e19da93c20b7be4f18d2de6e13)
If
, then
exists
![{\displaystyle c=\Gamma ^{-1}d\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c893e3d375270983c43557ad09fd8dc4561be3ff)
thus,
exists
![{\displaystyle d={\begin{bmatrix}d_{1}\\d_{2}\end{bmatrix}}={\begin{bmatrix}(2)(4)+(7)(2)\\(1.5)(4)+(3)(2)\end{bmatrix}}={\begin{bmatrix}22\\12\end{bmatrix}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8ce3752f72386298752535e4b1bc06a90461405)
![{\displaystyle {\begin{bmatrix}c_{1}\\c_{2}\end{bmatrix}}={\begin{bmatrix}53&24\\24&11.25\end{bmatrix}}^{-1}{\begin{bmatrix}22\\12\end{bmatrix}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0252606c558fd821b8457659c4c4e59837b90a14)
![{\displaystyle {\begin{bmatrix}c_{1}\\c_{2}\end{bmatrix}}={\begin{bmatrix}-2\\5.33\end{bmatrix}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d53e8ee2038a1594b69205c9f0f1cf76ab732b5)
![{\displaystyle v=4e_{1}+2e_{2}\equiv c_{1}b_{1}+c_{2}b_{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89978da9d9c09c390c2c7daada3afff1ab87e9e1)
![{\displaystyle c_{1}b_{1}+c_{2}b_{2}=(-2)(2e_{1}+7e_{2})+(5.33)(1.5e_{1}+3e_{2})\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/654e6f803e250744d6c474382c48c1226558aedd)
![{\displaystyle c_{1}b_{1}+c_{2}b_{2}=-4e_{1}-14e_{2}+8e_{1}+16e_{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41ff51deec2afb9c5fec52630d143a4f782a3b47)
![{\displaystyle c_{1}b_{1}+c_{2}b_{2}=4e_{1}+2e_{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c74765bbbb0c86ae30493747d5dda3a0c4cef75)
![{\displaystyle v=4e_{1}+2e_{2}\equiv c_{1}b_{1}+c_{2}b_{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89978da9d9c09c390c2c7daada3afff1ab87e9e1)
solution is correct
Find the integral
for
and ![{\displaystyle n=1\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68f5476359113900ec89cdda3a9ae9b03c80c4f6)
Using integration by parts, and then with the help of of
General Binomial Theorem
![{\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{n-k}y^{k}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb9e01c4b990d4f80eaa8d8f9b26fabc652ca912)
For
:
For substitution by parts,
![{\displaystyle \int log(1+x)dx=xlog(1+x)-\int (1-{\frac {1}{1+x}})dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4aff68e12cc3fa585bdef420da23a9d4666b6ca2)
![{\displaystyle \int log(1+x)dx=xlog(1+x)-x+log(1+x)+C\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6cefebf891160381a44f89103c6227483652a1a)
Therefore:
Using the General Binomial Theorem:
![{\displaystyle (x+y)^{0}=\sum _{k=0}^{0}{\binom {0}{k}}x^{0-k}y^{k}=1\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b864a9daf3877eb8ad3f6145350235c479cc8bc)
Therefore:
Which we have previously found that answer as:
For
:
Initially we use the following substitutions:
First let us consider the first term: ![{\displaystyle \int tlog(t)dt\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/325b2b73d5ce5b8282717fa4e6114537446f7bd5)
Next, we use the integration by parts:
![{\displaystyle \int tlog(t)dt={\frac {1}{2}}t^{2}log(t)-\int {\frac {1}{2}}t^{2}({\frac {1}{t}}dt)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/815b8af76613153f28025d8be6566832b45f47c6)
![{\displaystyle \int tlog(t)dt={\frac {1}{2}}t^{2}log(t)-\int {\frac {1}{2}}tdt)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9235473fbdc2b4dd04a81ea918299f20a33f46a)
![{\displaystyle \int tlog(t)dt={\frac {1}{2}}t^{2}log(t)-{\frac {1}{4}}t^{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bbdc0768a7b12a53951a21965def061f856f3cef)
Next let us consider the second term: ![{\displaystyle \int log(t)dt\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb77bc6b5b8ab49f73422528d58a154ec8bfa7f9)
Again, we will use integration by parts:
![{\displaystyle \int tlog(t)dt=tlog(t)-\int t({\frac {1}{t}}dt)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/368a0ec231066eb8e9a95f33ab963555031ab92a)
![{\displaystyle \int tlog(t)dt=tlog(t)-\int dt\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b105d4d72517668b2c117c0aa487f5c468f569f0)
![{\displaystyle \int tlog(t)dt=tlog(t)-t\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5087880aff91b247330cd338f1215a3d5001bc2)
Therefore:
![{\displaystyle \int (tlog(t)-\log(t))dt={\frac {1}{2}}t^{2}log(t)-{\frac {1}{4}}t^{2}-(tlog(t)-t)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88e5d9d6bff19ee9c89b87e7385b185caef33b61)
![{\displaystyle \int (tlog(t)-\log(t))dt={\frac {1}{2}}t^{2}log(t)-{\frac {1}{4}}t^{2}-tlog(t)+t\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dcd49930e7b7e6d3b6707525281dc1dce64190a9)
Re-substituting for t:
![{\displaystyle \int xlog(1+x)dx={\frac {1}{2}}(1+x)^{2}log(1+x)-{\frac {1}{4}}(1+x)^{2}-(1+x)log(1+x)+(1+x)+C\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eaac8f2dbec5b73a7eb82f4233916175c685eab7)
![{\displaystyle \int xlog(1+x)dx=(1+x)({\frac {1}{2}}(1+x)log(1+x)-{\frac {1}{4}}(1+x)-log(1+x)+1)+C\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ccc74d27f1df3b06a23f2edaea2436f2a301fbb0)
![{\displaystyle \int xlog(1+x)dx=(1+x)({\frac {1}{2}}xlog(1+x)-{\frac {1}{2}}log(1+x)-{\frac {1}{4}}x+{\frac {3}{4}})+C\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4dba911d163123a31b5cbb4d66f4070f4f4b3d07)
Therefore:
Using the General Binomial Theorem for the integral with t substitution
:
![{\displaystyle (x+y)^{1}=(t+(-1))^{1}=(t+(-1))=\sum _{k=0}^{1}{\binom {1}{k}}x^{1-k}y^{k}=\sum _{k=0}^{1}{\binom {1}{k}}t^{1-k}(-1)^{k}=t-1\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/025b748816d4bb31e3971506d664dde42d4d1b4e)
Therefore:
Which we have previously found that answer as:
Solved by: Gonzalo Perez
Consider the L2-ODE-CC (5) p.7b-7 with
as excitation:
(5) p.7b-7
(1) p.7c-28
and the initial conditions
.
Project the excitation
on the polynomial basis
(1)
i.e., find
such that
(2)
for x in
, and for n = 3, 6, 9.
Plot
and
to show uniform approximation and convergence.
Note that:
(3)
To solve this problem, it is important to know that the scalar product is defined as the following:
.
Therefore, it follows that:
, where
and
.
We know that if
are linearly independent, then by theorem on p.7c-37, the matrix is solvable.
According to this and (3)p.8-14:
If
exists
. (3)p.8-14
Now let's define the Gram matrix
as a function of
:
(1)p.8-13
Defining the "d" matrix as was done in (3)p.8-13, we get:
. (3)p.8-13
And according to (1)p.8-15:
(1)p.8-15
Now, we can find the values to compare
to
.
Using Matlab, this is the code that was used to produce the results:
The Matlab code above produced the following graph:
Where
is represented by the dashed line and the approximation,
, is represented by the red line. This code can work for all n values.
In a seperate series of plots, compare the approximation of the function
by Taylor series expansion about
.
Where:
For n=1:
For n=2:
For n=3:
For n=4:
For n=5:
For n=6:
For n=7:
For n=8:
For n=9:
For n=10:
For n=11:
For n=12:
For n=13:
For n=14:
For n=15:
For n=16:
Using Matlab to plot the graph:
Find
such that:
(1) p.7c-27
with the same initial conditions as in (2) p.7c-28.
Plot
for n = 3, 6, 9, for x in
.
In a series of separate plots, compare the results obtained with the projected excitation on polynomial basis to those with truncated Taylor series of the excitation. Plot also the numerical solution as a baseline for comparison.
First, we find the homogeneous solution to the ODE:
The characteristic equation is:
![{\displaystyle (\lambda -2)(\lambda -1)=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af5d3775e3da754de856d4bda6f0a85f8b327555)
Then,
![{\displaystyle \lambda =1,2\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/665b8a8e6b8fcd57dc28291fa259644ff6ba659a)
Therefore the homogeneous solution is:
![{\displaystyle y_{h}=C_{1}e^{(}2x)+c_{2}e^{x}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0903940be1db8899b50568a3a9924c71b73d2481)
Now to find the particulate solution
For n=3:
We can then use a matrix to organize the known coefficients:
Then, using MATLAB and the backlash operator we can solve for these unknowns:
![](//upload.wikimedia.org/wikiversity/en/3/3d/4_3n4.PNG)
Therefore
![{\displaystyle y_{p4}=4.0444+3.7458x+1.5743x^{2}+0.3981x^{3}+0.0543x^{4}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a89d48a5e2943c12c7ff84b38b557bd611bab7de)
Superposing the homogeneous and particulate solution we get
![{\displaystyle y_{n}=4.0444+3.7458x+1.5743x^{2}+0.3981x^{3}+0.0543x^{4}+C_{1}e^{2x}+C_{2}e^{x}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9f28bf5ba855dac54eb8f80cd3039ab09cbb742d)
Differentiating:
Evaluating at the initial conditions:
![{\displaystyle y(-0.75)=0.9698261719+0.231301601C_{!}+0.4723665527C_{2}=1\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e32e527292d31e950f9235e32fed6d2ed806f67)
![{\displaystyle y'(-0.75)=1.9645125+0.4462603203C_{1}+0.4723665527C_{2}=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/156b393b4d9061f8567100ab618899d967c5d691)
We obtain:
![{\displaystyle C_{1}=-4.46\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aedbb96c22531f7d5bfad39b43a6e3d57b694c20)
![{\displaystyle C_{2}=0.055\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1b952e8e3f092bf1a8f336333a1f79661218cec)
Finally we have:
![{\displaystyle y_{n}=4.0444+3.7458x+1.5743x^{2}+0.3981x^{3}+0.0543x^{4}-4.46e^{2x}+0.055e^{x}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8dbfda5a566057849a420cc69b0500d5109c61bc)
For n=6:
We can then use a matrix to organize the known coefficients:
Then, using MATLAB and the backlash operator we can solve for these unknowns:
![](//upload.wikimedia.org/wikiversity/en/8/88/4_3n7.PNG)
Therefore
![{\displaystyle y_{p7}=377.4833+375.3933x+185.6066x^{2}+60.5479x^{3}+14.4946x^{4}+2.6492x^{5}+0.3619x^{6}+0.0310x^{7}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0db9b140684a1139aa91da23c7819c56ce9cf06a)
Superposing the homogeneous and particulate solution we get
![{\displaystyle y_{n}=377.4833+375.3933x+185.6066x^{2}+60.5479x^{3}+14.4946x^{4}+2.6492x^{5}+0.3619x^{6}+0.0310x^{7}+C_{1}e^{2x}+c_{2}e^{x}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7c6a2cb5d4a0fe0f16542800ee2c54820878f8c)
Differentiating:
Evaluating at the initial conditions:
![{\displaystyle y(-0.75)=178.816+0.2231301601C_{!}+0.4723665527C_{2}=1\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8bd60f405c19cb1867ad39fea28fcae0340f7277)
![{\displaystyle y'(-0.75)=178.413+0.4462603203C_{1}+0.4723665527C_{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a809fa359950e34c98c6ffd76ed67fbe95979e91)
We obtain:
![{\displaystyle C_{1}=-2.6757\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83757e6fc287a997666578f9a5f5e82e2827ae64)
![{\displaystyle C_{2}=-375.173\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a44af748b654cb29dbd44c423dbfd08a02acb5d)
Finally
![{\displaystyle y_{n}=377.4833+375.3933x+185.6066x^{2}+60.5479x^{3}+14.4946x^{4}+2.6492x^{5}+0.3619x^{6}+0.0310x^{7}+-2.6757e^{2x}-375.173e^{x}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5257dfd573f2e9d368a2b276c6a1af5e9fd3a4e9)
For n=9:
![{\displaystyle r(x)=\sum _{0}^{n}-{\frac {(-1)^{n}x^{n}}{n\ln(10)}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1995083561ffc8842cfab6eef6aba7e3e9281a5c)
We can then use a matrix to organize the known coefficients:
Then, using MATLAB and the backlash operator we can solve for these unknowns:
![](//upload.wikimedia.org/wikiversity/en/2/21/4_3n11.png)
Therefore
![{\displaystyle y_{p11}=1753158.594+1752673.419x+875851.535x^{2}+291627.134x^{3}+72745.1129x^{4}+14484.362x^{5}+2392.510x^{6}+335.632x^{7}+40.417x^{8}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec8ba268d7198cab32108791a23c66dee9dc5cbb)
![{\displaystyle +4.1499x^{9}+0.3474x^{(}10)+0.0197x^{(}11)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92463a4196e019b60b2df7e3785d3b4bc7b8cb01)
Superposing the homogeneous and particulate solution we get
![{\displaystyle y_{n}=1753158.594+1752673.419x+875851.535x^{2}+291627.134x^{3}+72745.1129x^{4}+14484.362x^{5}+2392.510x^{6}+335.632x^{7}+40.417x^{8}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f86722e35ebd573cf3c5f3d4b230f0284c5c986e)
![{\displaystyle +4.1499x^{9}+0.3474x^{(}10)+0.0197x^{(}11)+C_{1}e^{2x}+C_{2}e^{(}x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93d03a30bbf5f2cb2b24deee6a0e43bbc66774ff)
Differentiating:
![{\displaystyle +1.7517x10^{6}x+1.75267x10^{6}+2C_{1}e^{2x}+C_{2}e^{x}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a46d216922a46764f3fa5294eb3ec056f41e4bc6)
Evaluating at the initial conditions:
![{\displaystyle y(-0.75)=828254+0.2231301601C_{!}+0.4723665527C_{2}=1\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10a35fb9b8815d776be1dda670f731032d4db46a)
![{\displaystyle y'(-0.75)=828145+0.4462603203C_{1}+0.4723665527C_{2}=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2511d8a7f7027a9a1bf34d33ec1ad391197277f)
We obtain:
![{\displaystyle C_{1}=-484.022\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d67cf176b9204692655649db5499ce3ebc5d58fc)
![{\displaystyle C_{2}=-1753750\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3a77d32b8ebe63a3a5a5c547bf00547165ef6ae)
Finally
![{\displaystyle y_{n}=1753158.594+1752673.419x+875851.535x^{2}+291627.134x^{3}+72745.1129x^{4}+14484.362x^{5}+2392.510x^{6}+335.632x^{7}+40.417x^{8}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f86722e35ebd573cf3c5f3d4b230f0284c5c986e)
![{\displaystyle +4.1499x^{9}+0.3474x^{(}10)+0.0197x^{(}11)-484.022e^{2x}-1753750e^{x}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33da837d3ac3e0d70f9d034e893045c45d4e4b3e)
Here is the graph for this problem using Matlab: