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.

Tutors, sign in to answer this question.

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.

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.