Answer:
[tex](x + 2x^2)-\frac{1}{18}\ln |15+18y|=C'[/tex]
Step-by-step explanation:
Given differential equation,
[tex]\frac{dy}{dx}=15+30x+18y + 36xy[/tex]
[tex]\frac{dy}{dx}=15(1+2x)+18y(1+2x)[/tex]
[tex]\frac{dy}{dx}=(15+18y)(1+2x)[/tex]
[tex]\frac{dy}{15+18y}=(1+2x)dx------(1)[/tex]
Let 15 + 18y = t
18dy = dt
[tex]\implies dy=\frac{dt}{18}[/tex]
From equation (1),
[tex]\frac{1}{18} \frac{1}{t}dt = (1+2x)dx[/tex]
Integrating both sides,
[tex]\frac{1}{18}\ln |t| = x + 2x^2 + C[/tex]
[tex]\frac{1}{18}\ln |15+18y| = x + 2x^2 + C[/tex]
[tex]\implies (x + 2x^2)-\frac{1}{18}\ln |15+18y|=C'[/tex] ( where C' = -C = constant)
Which is the required equation,
Where,
[tex]G(x) = x+2x^2\text{ and }H(y) = -\frac{1}{18}\ln |15+18y|[/tex]
