Answer to Question #159388 in Differential Equations for Leila

Question

(xz+yz)dz/dy + (xz-yz)dz/dy = x²+y² find the general solution

Answer 1

$(xz(y)+yz(y))z'(y)+(xz(y) - yz(y))z'(y)=x^2+y^2$

Ordinary differential equation classification:

first order nonlinear ordinary differential equation

$\frac{\large dz(y)}{\large dy}(xz(y)-yz(y))+\frac{\large dz(y)}{\large dy}(xz(y)+yz(y))=x^2+y^2$

Solve for $\frac{\large dz(y)}{\large dy}$

$\space\space\frac{\large dz(y)}{\large dy} = \frac{\large x^2+y^2}{\large 2xz(y)}$

Multiply both sides by $2xz(y):$

$2x\frac{\large dz(y)}{\large dy}z(y)=x^2+y^2$

Integrate both sides with respect to y:

$\int 2x\frac{\large dz(y)}{\large dy}z(y)dy=\int (x^2+y^2)dy$

Evaluate the integrals:

$xz(y)^2 = \frac{\large y^3}{\large 3}+x^2y+C_1.$       where $C_1$ is an arbitrary constant.

Solve for z(y):

$Answer: \space z(y) = - \large\frac{\sqrt{y^3+3x^2y+3C_1}}{\sqrt{3x}} \space or \space \large\frac{\sqrt{y^3+3x^2y+3C_1}}{\sqrt{3x}}$