Question

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

Answer 1

Answer:

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

Step-by-step explanation:

Given that

$\dfrac{dy}{dt}=\dfrac{2t}{y+yt^2}$

This is a differential equation.

Now by separating variables

$y dy= \dfrac{2t}{1+t^2}dt$

Now by integrating both side

$\int y dy=\int \frac{2t}{1+t^2}dt$

Now by soling above integration

We know that  integration of dx/x is lnx.

$\dfrac{y^2}{2}=\ln (1+t^2)+C$

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  $y^2=2\ln (1+t^2)+4$