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2 Answers

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


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/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.

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