Question

2yy′=x  and  y(2)=3.

Answer 1

2yy' = x

=> $2y\frac{dy}{dx}=x$

=> 2ydy = xdx

=> $\int2ydy = \int xdx$ + C' , where C' is integration constant.

=> $2\int ydy = \int xdx$ + C'

=> $2\frac{y²}{2}=\frac{x²}{2}$ + C'

=> 2y² = x² + 2C'

=> 2y² = x² + C Let C = 2C'

So the general solution of the given differential equation is 2y² = x² + C where C is integration constant.

Now given that y(2)=3

i.e. when x =2 , y = 3

So 2*3² = 2² + C

=> C = 18-4 = 14

So the particular solution of the given differential equation is 2y² = x² + 14