# Solve the IVP (x \ln (y^4) + 4 \ln y - y^2) dx + \left ( {-2y + \frac{4x}{y}} \right ) dy = 0, ...

## Question:

Solve the IVP

{eq}(x \ln (y^4) + 4 \ln y - y^2) dx + \left ( {-2y + \frac{4x}{y}} \right ) dy = 0, \; y(0) = 1 {/eq}.

## Solving a Nonexact Differential Equation:

In order to solve the differential equation {eq}M(x, y) \: dx + N(x, y) \: dy = 0, {/eq} we first check to see if it is exact. An exact differential equation satisfies the condition {eq}\displaystyle\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. {/eq} If the condition is not satisfied but {eq}\displaystyle\frac{M_y - N_x}{N} {/eq} is a function of {eq}x {/eq} alone, then the function {eq}\mu(x) = e^{\int \frac{M_y - N_x}{N}} \: dx {/eq} is an integrating factor for the differential equation. After multiplying the equation by the integrating factor, the new equation will be exact.

Become a Study.com member to unlock this answer!

Given the initial value problem

{eq}(x \ln (y^4) + 4 \ln y - y^2) dx + \left ( {-2y + \displaystyle\frac{4x}{y}} \right ) dy = 0, \; y(0) = 1 {/eq},...