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Find the general solution?

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. 

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Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)

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

Xavier J. | Tutor in Math, topics range from Algebra to Calculus.Tutor in Math, topics range from Algebra...
5.0 5.0 (3 lesson ratings) (3)

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.