Find the general solution?

Question

Find the general solution of ty'-y=t^2*e^(-t), t>0.

 

I got y=(t^-1)(e^-t)(t^3/3+c) but when I simplify that I didn't get the right answer. Please help me with steps. 

Answer 1

This is a typical first-order linear differential equation: y' + Py = Q

Here is a standard way to solve it.

Multiply both sides of y' + Py = Q by e^∫P,

e^∫P y' + e^∫P Py = e^∫PQ

Apply product rule,

(e^∫P y)' = e^∫P Q

Solve for y,

y = (1/e^∫P)(∫e^∫P Q dx + c) <==Formula of the solution for any first-order linear differential equation

For your problem: ty'-y=t^2*e^(-t), t>0.

y' - y/t = t e^(-t)

p = -1/t

∫P = -lnt

e^∫P = 1/t

So,

y = t[∫e^(-t) dt + c ]

= t[-e^(-t) + c] <==Answer

Answer 2

1st thing we do in this problem is put it in a form that will allow us to find an integrating factor. We do this by multiplying the entire equation by (1/t), this gives us y'-(1/t)y=te^-t

From here we see that our integrating factor is g(t)=e^-∫(1/t)= 1/t.

Multiplying the equation by the integrating factor we get  (1/t)y'-(1/t²)y=e^-t

This then simplifies to  ((1/t)y)'=e^-t

Integrate both sides then solve for y and you should get y(t)=-te^-t

This is only the non homogeneous part of the solution, the homogeneous solution would be y(t)=ct.

Please let me know if need me to elaborate more.