Question

Solve the IVP dy/dt = 2t/(y + t^2 y), y(0) = -2.

Answer 1

`y^2=2\ln (1+t^2)+4`

Explanation:

Given that

`(dy)/(dt)=(2t)/(y+yt^2)`

This is a differential equation.

Now by separating variables

`y dy= (2t)/(1+t^2)dt`

Now by integrating both side

`\int y dy=\int (2t)/(1+t^2)dt`

Now by soling above integration

We know that integration of dx/x is lnx.

`(y^2)/(2)=\ln (1+t^2)+C`

Where C is the constant.

`y^2=2\ln (1+t^2)+C`

Given that when t=0 then y= -2

So by putting the above values of t and y we will find C

4=2 ln(1)+C (we know that ln(1)=0)

So C=4

⇒
`y^2=2\ln (1+t^2)+4`

So solution of above equation is
`y^2=2\ln (1+t^2)+4`