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.
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.
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^{∫P}Q
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
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/t^{2})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.