Find the solution for the following exact differential equations.

Question

(2ty/t^2+1) - 2t - (2 - ln(t^2 + 1))y' = 0 ; y(5) = 0   Note: the "+1" is not part of the exponent in "(2ty/t^2+1)"

 

My answer to this problem is "y(ln(t^2+1)-2) = 0" Am I right?

Answer 1

((2ty/t²+1) - 2t)dt - (2 - ln(t² + 1))dy = 0 ; y(5) = 0

F(t, y)=(2+ln(t² + 1))y-t²

(I guess you know where F(t, y) is coming from)

F_t(t, y)=2ty/(t²+1)-2t

Fy(t, y)=(2+ln(t2 + 1))

So, the general solution is F(t, y)=C

F(5, 0)=-25=C

(2+ln(t² + 1))y-t²+25=0.

You are slightly wrong.

Check this solution.