# Determine if the ODE is exact. If it is exact, solve it a) (2xy - y^2) + (2xy - x^2)y^' = 0 3x^2...

## Question:

Determine if the ODE is exact. If it is exact, solve it.

a) {eq}(2xy - y^2) + (2xy - x^2){y}' = 0 {/eq}

b) {eq}(3x^2 - 2xy + 2)dx + (6y^2 - x^2 + 3)dy = 0 {/eq}

c) {eq}(e^xsiny - 2y sinx) + (e^xcosy + 2cosx){y}' = 0 {/eq}

d) {eq}(t \ln y + ty) + (y \ln t + ty)\frac{dy}{dt} = 0, t > 0, y > 0{/eq}

## Exact differential equation:

A differential equation of the form {eq}M(x,y)dx + N(x,y)dy=0 {/eq} is said to be exact if {eq}\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}. {/eq}

The solution of the exact differential equation is given by {eq}\int M(x,y)dx +\int N_{y} dy=0 {/eq}

where {eq}N_y {/eq} indicates, those terms in {eq}N(x,y) {/eq} which are free from {eq}x {/eq}

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a) The given differential equation is

{eq}(2xy - y^2) + (2xy - x^2){y}' = 0 \\ (2xy - y^2) + (2xy - x^2)\dfrac{dy}{dx} = 0 \\ (2xy - y^2) dx + (2xy...

First-Order Linear Differential Equations

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Chapter 16 / Lesson 3
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In this lesson you'll learn how to solve a first-order linear differential equation. We first define what such an equation is, and then we give the algorithm for solving one of that form. Specific examples follow the more general description of the method.