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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.

Answer and Explanation: 1

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Given equation is

{eq}\displaystyle \left(\left(1/x\right)\:+\:2xy^2\right)dx+\left(2x^2y-cos\:y\right)dy=0, \ y(1) = \pi {/eq}.

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First-Order Linear Differential Equations

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Chapter 16 / Lesson 3
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Learn to define what a linear differential equation and a first-order linear equation are. Learn how to solve the linear differential equation. See examples.


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