Solve the following by the integrating factor method. (4y/t)+y'=t^(-3) and y(1)=2?

Question

Solve the following ordinary differential equation by the integrating factor method. (4y/t)+y'=t^(-3) and y(1)=2.
This equation can be written in the form y'+f(t)y=g(t), when this happens, one should calculate the integration factor.
1) What is the integrating factor in this case? (just enter an expression)
2) Enter an expression for the antiderivative
3) What is the solution to the initial value problem? (Enter an expression for y(t), not an equation)

Answer 1

1) t^4
2) 1/2t^2+3/2
3) 1/(2*t^2)+3/(2*t^4)

1) e^ (Integral of 4/t) = e^(4log(t)) = t^4
2) equation * t^4 becomes t^4*y'+4t^3*y = t so [t^4*y]' = t you thenintegrate this to t^4*y=1/2t^2+C and substitute t=1 and y=2 into this to get 2=1/2+C and solve for C=3/2
3) finally solve for y by dividing by t^4

Answer 2

properly this differential equation is of the type dy/dx+Py=Q the place P=2tanx and Q=sinx. to sparkling up this equation first you will might desire to calculate the integrating ingredient which would be equivalent to integration of 2tanx dx. after looking the integrating ingredient positioned the values interior the equation y+I.F=INTEGRATION OF sinx.I.F dx. and you will get the answer.