Solve the Initial Value Problem: y' = -(sinx)(y) + (x)(e^(cosx)); y(0) = 1


Solve the Initial Value Problem: {eq}\displaystyle y{}'=-\sin x(y)+xe^{\cos x} {/eq}; {eq}\displaystyle y(0)=1 {/eq}

Solution of the Differential Equation:

The given differential equation is first order linear differential equation. We compare the given differential equation with the differential equation {eq}\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}+P(x)y=Q(x) {/eq} and find the values of the function {eq}P(x) {/eq} and {eq}Q(x) {/eq} which is function of {eq}x. {/eq} To find the solution of the differential equation we find the integrating factor of the differential equation and then the solution of the differential equation will be written in the form of

{eq}\displaystyle y\times \textbf{I.F}=\int Q(t)\times \textbf{I.F}dt+c {/eq}

Where, {eq}c {/eq} is a constant of integration.

Answer and Explanation:

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Consider the differential equation

{eq}\displaystyle y{}'=-\sin x(y)+xe^{\cos x},\quad y(0)=1\\ \displaystyle \frac{dy}{dx} + \sin x\: y = xe^{\cos...

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First-Order Linear Differential Equations


Chapter 16 / Lesson 3

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.

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