Question

3. Find the general solution of the following differential equation. 1 (1 += sin²z) de +(+²2) dy-0 PROBLEM STEP BY STEP 2y EXPLAIN IN WORDS HOW TO SOLVE THIS 3. Find the general solution of the following differential equation . 1 ( 1 + = sin²z ) de + ( + ²2 ) dy - 0 PROBLEM STEP BY STEP 2y EXPLAIN IN WORDS HOW TO SOLVE THIS​

Answer 1

Multiply both sides by $y$ to get an exact equation.

$\left(1 + \dfrac{x^3}y \sin^2(x)\right) \, dx + \dfrac1y \left(x + \dfrac1{\cos^2(2y)}\right) \, dy = 0$

$\implies \underbrace{(y + x^3 \sin^2(x))}_{M(x,y)} \, dx + \underbrace{(x + \sec^2(2y))}_{N(x,y)} \, dy = 0$

This ODE is exact since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Then the solution is given by an implicit function

$f(x,y) = C$

Taking differentials on both sides by the chain rule gives

$\dfrac{\partial f}{\partial x} \, dx + \dfrac{\partial f}{\partial y} \, dy = 0$

so that we have the system of partial differential equations

$\dfrac{\partial f}{\partial x} = M = y + x^3 \sin^2(x)$

$\dfrac{\partial f}{\partial y} = N = x + \sec^2(2y)$

Integrate both sides of the first of these equations with respect to $x$ to recover $f$.

$\displaystyle \int \frac{\partial f}{\partial x} \, dx = \int (y + x^3 \sin^2(x)) \, dx$

$\implies f(x,y) = xy + g(x) + h(y)$

where $g(x)$ is the antiderivative of $x^3\sin^2(x)$ (and is easy enough to compute by parts).

Differentiating both sides with respect to $y$ gives

$\dfrac{\partial f}{\partial y} = x + \dfrac{dh}{dy} = x + \sec^2(2y)$

$\implies \dfrac{dh}{dy} = \sec^2(2y)$

$\implies h(y) = \dfrac12 \tan(2y) + C$

Then the general solution to the ODE is

$f(x,y) = \boxed{xy + g(x) + \dfrac12 \tan(2y) = C}$