Question

Solve the initial value problem ty'+3y=0 , y(1)=2 t>0

Answer 1

`y=(2)/(t^3)`

Explanation:

Transforming the equation:

`ty'+3y=0\\\\ty'=-3y\\\\t((dy)/(dt))=-3y\\ \\tdy=-3ydt\\\\(dy)/(y) =-(3dt)/(t)`

Solving the separable equation:

`\int(1)/(y) dy=\int-(3)/(t) dt\\\\ln(y)=-3ln(t)+c\\\\y=(e^(c) )/(t^(3) )`

Finding c:

since y(1)=2 we get:

`2=(e^c)/(1^(3) ) \\\\c=ln(2)`

Final solution:

`y=(e^(^l^n^(^2^)^))/(t^3)\\\\y=(2)/(t^3)`