# Solve the IVP {eq}(1/x) + 2xy^2) \ dx + (2x^2y - \cos y) \ dy = 0, \ y(1) = \pi {/eq}.

## Question:

Solve the IVP {eq}(1/x) + 2xy^2) \ dx + (2x^2y - \cos y) \ dy = 0, \ y(1) = \pi {/eq}.

## Solve the Initial Value Problem.

The differential equation of type {eq}\displaystyle M\:dx+N\:dy=0 {/eq} is called exact differential equation when it satisfies the condition {eq}\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} {/eq}.

Then solution will be written as

{eq}\displaystyle \int _{y\:constant}\:M\:dx+\int _{NO\:x\:term}\:N\:dy=C {/eq}.

Use the initial conditions to find the constant values.