Question

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

\((y + \sin y){y^\prime } = x + {x^3}\)

Short answer

The solution is \(\frac{{{y^2}}}{2} - \cos y = \frac{{{x^2}}}{2} + \frac{{{x^4}}}{4} + C\).

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\((y + \sin y){y^\prime } = x + {x^3}\).

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

\(\begin{aligned}(y + \sin y)\frac{{dy}}{{dx}} &= x + {x^3}\\(y + \sin y)dy &= \left( {x + {x^3}} \right)dx\end{aligned}\)

Integrate

Integrating both sides we get:

\(\begin{aligned}\int {(y + \sin y)} dy &= \int {\left( {x + {x^3}} \right)} dx + C\\\int y dy + \int {\sin } ydy &= \int x dx + \int {{x^3}} dx + C\\\frac{{{y^{1 + 1}}}}{{1 + 1}} - \cos y &= \frac{{{x^{1 + 1}}}}{{1 + 1}} + \frac{{{x^{3 + 1}}}}{{3 + 1}} + C {\rm{ Using formulas }}\int {{x^n}} dx &= \frac{{{x^{n + 1}}}}{{n + 1}},n \ne - 1,\int {\sin } xdx &= - \cos x\\\frac{{{y^2}}}{2} - \cos y &= \frac{{{x^2}}}{2} + \frac{{{x^4}}}{4} + C\end{aligned}\)

Now equation can’t be solve further.