Question

Solve the following differential equation:

(7x+8y)dx+(8x-2)dy=0

Answer 1

Since $\frac{\partial(7x+8y)}{\partial y}=8=\frac{\partial(8x-2)}{\partial x},$ we conclude that the equation is exact, and hence there exists the function $u=u(x,y)$ such that

$\frac{\partial u}{\partial x}=7x+8y, \ \frac{\partial u}{\partial y}=8x-2.$

It follows that $u=\frac{7}2x^2+8xy+c(y).$

Then $\frac{\partial u}{\partial y}=8x+c'(y)=8x-2,$ and hence $c'(y)=-2.$ Therefore, $c(y)=-2y+C.$

We conclude that the general solution of the differential equation is of the form:

$\frac{7}2x^2+8xy-2y=C.$