1. ## nonhomogenous equations

Consider the differential equation ay''+by'+cy=d where d is a constant. Show that every solution of the equation approaches d/c as t goes to infinity. What happens if c=0? What if b=0?

Setting Y(t)=A, and following from that Y'(t)=0 and Y''(t)=0. I plugged these values in to the differential equation and got that A=d/c but I confused about how to find the homogenous solution and how do we know that that will go to 0 as t goes to infinity.

thanks!

2. Originally Posted by morganfor
Consider the differential equation ay''+by'+cy=d where d is a constant. Show that every solution of the equation approaches d/c as t goes to infinity. What happens if c=0? What if b=0?

Setting Y(t)=A, and following from that Y'(t)=0 and Y''(t)=0. I plugged these values in to the differential equation and got that A=d/c but I confused about how to find the homogenous solution and how do we know that that will go to 0 as t goes to infinity.

thanks!
I don't think it is true unless you restrict at lease some of the values of a,b and c.

To solve the complimentry solution use the ansatz $y=e^{rt} \implies y'=re^{rt} \implies y''=r^2e^{rt}$

Then plut into the homogenous equation to get

$ay''+by'+cy=0 \iff ar^2e^{rt}+bre^{rt}+ce^{rt}=0$

factoring gives

$e^{rt}(ar^2+br+c)=0$

Since the exponential function is never zero the quadratic must be zero.
Using the quadratic equation we gt

$r=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

So the complimetery solution to the equation is

$y_c=c_1e^{ \frac{-b + \sqrt{b^2-4ac}}{2a}t}+c_2e^{\frac{-b- \sqrt{b^2-4ac}}{2a}t}$

This will not always tend to zero as t goes to infinity. I.e if $b,c > 0 \text{ and } a < 0$

The solution of the homoenous problem goes to infinity.

I hope this helps