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=\frac1{17},b_0=0,b_1=-\frac3{128},b_2=0,b_3=\frac1{16}`

so that the particular solution is


`y_p=\frac1{17}e^x+\frac1{16}x^3-\frac3{128}x`

and the general solution to the ODE is


`y=y_c+y_p`

`y=C_1\cos4x+C_2\sin4x+\frac1{17}e^x+\frac1{16}x^3-\frac3{128}x`

Given the initial values, we have


`y(0)=4\implies4=C_1+\frac1{17}\implies C_1=(67)/(17)`

`y'(0)=0\implies0=4C_2+\frac1{17}-\frac3{128}\implies C_2=-(77)/(8704)`

so that the solution to the IVP is


`y=(67)/(17)\cos4x-(77)/(8704)\sin4x+\frac1{17}e^x+\frac1{16}x^3-\frac3{128}x`