Question

Solve the initial-value problem using the method of undetermined coefficients. y"+16y= e^x + x^3.
y(0)=4,
y'(0)=0

Answer 1

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

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