Question

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

\(x{y^2}y' = x + 1\)

Short answer

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

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

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

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

Integrate

Integrating both sides we get:

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

\(\frac{{{y^{2 + 1}}}}{{2 + 1}} = \frac{{{x^{0 + 1}}}}{{0 + 1}} + \ln |x| + C\;\;\;\)Using formulas\(\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C,n \ne - 1;\int {\frac{1}{x}} dx = \ln |x| + C\)

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

Simplify solution

Solving for\(y\)we have:

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

Where\(K = 3C\), and\(K\) is an arbitrary constant, since\(C\)is an arbitrary constant. Therefore, the solution of given the differential equation\(x{y^2}{y^\prime } = x + 1\) is\(y = \sqrt(3){{3x + 3\ln |x| + K}}\).