Solve the initial value problem 9(t+1)dy/dt - 5y = 20t for t>-1 with y(0) = 11 P

Question

Solve the initial value problem

9(t+1)dy/dt - 5y = 20t

for t>-1 with y(0) = 11

Put the problem in standard form.

Then find the integrating factor p(t) = ?

and finally find y(t) = ?

Explanation / Answer

9(t + 1) dy/dt - 5y = 20 t with condition t > - 1 we divide by 9(t + 1) : y' - 5y/[ 9(t + 1) ] = 20 t / 9(t + 1) Integrating factor : µ = e^?( -5 / ( 9(t + 1) ) ) dt = e^( ln(t + 1)^(-5/9) ) = (t + 1)^(-5/9) ( the condition t > -1 goes with the ln(t + 1) ) y' (t + 1)^(-5/9) - 5y(t + 1)^(- 14/9) / 9 = 20 t (t + 1)^(- 14/9) / 9 d(y (t + 1)^(-5/9) ) = 20 t (t + 1)^(- 14/9) / 9 dt ? d(y (t + 1)^(-5/9) ) = ? 20 t (t + 1)^(- 14/9) / 9 dt + C y (t + 1)^(-5/9) = ( 9 + 5t ) (t + 1)^(- 5/9) + C y = ( 9 + 5t ) + C(t + 1)^(5/9) y(0) = 9 + C = 14 so C = 5 y(t) = 9 + 5t + 5(t + 1)^(5/9)

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