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nexus9112 [7]
3 years ago
8

Solve the initial-value problem using the method of undetermined coefficients. y"+16y= e^x + x^3.

Mathematics
1 answer:
il63 [147K]3 years ago
6 0
The homogeneous part of the ODE has characteristic equation

r^2+16=0

with roots at r=\pm4i, so the characteristic solution would be

y_c=C_1\cos4x+C_2\sin4x

As a guess for the solution to the nonhomogeneous part, we can try

y_p=ae^x+b_0+b_1x+b_2x^2+b_3x^3
\implies{y_p}''=ae^x+2b_2+6b_3x

Substituting into the ODE gives

(ae^x+2b_2+6b_3x)+16(ae^x+b_0+b_1x+b_2x^2+b_3x^3)=e^x+x^3
17ae^x+(16b_0+2b_2)+(16b_1+6b_3)x+16b_2x^2+16b_3x^3=e^x+x^3
\implies\begin{cases}17a=1\\16b_0+2b_2=0\\16b_1+6b_3=0\\16b_2=0\\16b_3=1\end{cases}\implies a=\dfrac1{17},b_0=0,b_1=-\dfrac3{128},b_2=0,b_3=\dfrac1{16}

so that the particular solution is

y_p=\dfrac1{17}e^x+\dfrac1{16}x^3-\dfrac3{128}x

and the general solution to the ODE is

y=y_c+y_p
y=C_1\cos4x+C_2\sin4x+\dfrac1{17}e^x+\dfrac1{16}x^3-\dfrac3{128}x

Given the initial values, we have

y(0)=4\implies4=C_1+\dfrac1{17}\implies C_1=\dfrac{67}{17}
y'(0)=0\implies0=4C_2+\dfrac1{17}-\dfrac3{128}\implies C_2=-\dfrac{77}{8704}

so that the solution to the IVP is

y=\dfrac{67}{17}\cos4x-\dfrac{77}{8704}\sin4x+\dfrac1{17}e^x+\dfrac1{16}x^3-\dfrac3{128}x
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