Question

\({\bf{1 - 8}}\)Solve the differential equation.

\(\left( {{y^2} + x{y^2}} \right){y^\prime } = 1.\)

Short answer

The solution is\(y = \sqrt(3){{3\ln |x + 1| + K}},y = 0.{\rm{ }}\)

Step-by-step solution

Definition

A differential equation is an equation with one or more derivatives of a function.

Separate variables

Consider the differential equation\(\left( {{y^2} + x{y^2}} \right){y^\prime } = 1.\)

We know that\({y^\prime } = \frac{{dy}}{{dx}}\), so above equation can be written as:

\(\begin{aligned}\left({{y^2} + x{y^2}} \right)\frac{{dy}}{{dx}} &= 1\\{y^2}(x + 1)\frac{{dy}}{{dx}} &= 1\\{y^2}(x + 1)dy = dx\\{y^2}dy &= \frac{{dx}}{{x + 1}}\end{aligned}\)

Integrate

Integrating both sides we get:

\(\int {{y^2}} dy = \int {\frac{{dx}}{{x + 1}}} + C\)

\(\frac{{{y^{2 + 1}}}}{{2 + 1}} = \ln |x + 1| + C\;\)Using formulas \(\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C,n \ne - 1;\int {\frac{1}{x}} dx = \ln |x| + C\)

\(\frac{{{y^3}}}{3} = \ln |x + 1| + C\;\)

Simplify solution

Solving for\(y\)we have:

\(\begin{aligned}\frac{{{y^3}}}{3} &= \ln |x + 1| + C\\{y^3} &= 3\ln |x + 1| + 3C\\y &= \sqrt(3){{3\ln |x + 1| + 3C}}\\y &= \sqrt(3){{3\ln |x + 1| + K}}\end{aligned}\)

Also, \(y = 0\)is a solution.

Therefore, the solutions of given the differential equations\(\left( {{y^2} + x{y^2}} \right){y^\prime } = 1\) are\(y = \sqrt(3){{3\ln |x + 1| + K}},y = 0.{\rm{ }}\)