Find general solution to the non-homogeneous equation: $y'' - 16y' + 64y = \frac{e^{-8x}}{1 + x^2}$

Question

Find the most general solution to the non-homogeneous equation:

$$y'' - 16y' + 64y = \frac{e^{-8x}}{1 + x^2}$$

With the given conditions: $y(0) = -10$ and $y'(0) = 9$. I partially solved it, with

$$e^{8x} \int_0^x \frac{-te^{-16t}}{1+t^2}dt + xe^{8x} \int_0^x \frac{-e^{-16t}}{1+t^2}dt + Ce^{8x}+Dxe^{8x}$$

I am not sure what the variables C and D are. Can anyone help me find the variables?

Answer 1

OK, same guy here. I found out how to solve it:

$y"−16y′+64y=(e^{−8x})/(1+x^{2})$

And given the initial conditions y(0) = -10 and y'(0) = 9

To find C and D, we just make the left half of the equation equal 0 and solve it.

$y"−16y′+64y=0$

The general formula for the solution is:

$y = Ae^{8x}+Bxe^{8x}$ and $y' = 8Axe^{8x}+Ae^{8x}+Be^{8x}$

Applying the initial values into both formulas will give you:

A = 89 and B = -10

These coefficients are the same as D and C respectively.

Answer 2

Note $$ (ye^{-8x})''=(y''-16y'+64y)e^{-8x}$$ and hence the equation becomes $$ (ye^{-8x})''=\frac{e^{−16x}}{1+x^{2}}. $$ So $$ (ye^{-8x})'=\int_0^x\frac{e^{−16t}}{1+t^{2}}dt+C $$ and hence $$ ye^{-8x}+10=\int_0^x\int_0^s\frac{e^{−16t}}{1+t^{2}}dtds+Cx=\int_0^x\int_t^x\frac{e^{−16t}}{1+t^{2}}dsdt+Cx=\int_0^x\frac{(x-t)e^{−16t}}{1+t^{2}}dsdt+Cx. $$ So $$ y=e^{8x}\bigg[-10+\int_0^x\frac{(x-t)e^{−16t}}{1+t^{2}}dt+Cx\bigg]. $$ Now \begin{eqnarray} y'&=&8e^{8x}\bigg[-10+\int_0^x\frac{(x-t)e^{−16t}}{1+t^{2}}dt+Cx\bigg]\\ && +e^{8x}\bigg[\int_0^x\frac{e^{−16t}}{1+t^{2}}dt+C\bigg] \end{eqnarray} and using $y'(0)=9$, one has $$ 8(-10)+C=9 $$ which implies $C=89$. So the solution is $$ y=e^{8x}\bigg[-10+\int_0^x\frac{(x-t)e^{−16t}}{1+t^{2}}dt+89x\bigg]. $$