# Solve: ='false' \frac {dx}{dt}+6x=cost+sin5t

## Question:

Solve:

{eq}\frac {dx}{dt}+6x=cost+sin5t {/eq}

## First Order Linear Differential Equation:

A differential equation in the form of {eq}\displaystyle \frac{{dx}}{{dt}} + P(t)y = Q(t) {/eq}, is called the linear differential equation.

Where P(t) and Q(t) are function of t or constants.

General solution of linear differential equation is obtained by

{eq}\displaystyle x.IF = \int {IF.Qdt + K} {/eq}

Where IF is known as integrating factor and K is the constant of the integration.

Integrating factor is given by: {eq}\displaystyle IF = {e^{\int {Pdt} }} .{/eq}

Required formulas:

{eq}\displaystyle \eqalign{ & 1.\,\int {{e^{at}}\sin bt\,dt = \frac{{{e^{at}}}}{{{a^2} + {b^2}}}\left( {a\sin bt - b\cos bt} \right)} \cr & 2.\,\int {{e^{at}}\cos bt\,dt = \frac{{{e^{at}}}}{{{a^2} + {b^2}}}\left( {a\cos bt + b\sin bt} \right)} .\cr} {/eq}

Given differential equation: {eq}\displaystyle \frac{{dx}}{{dt}} + 6x = \cos t + \sin 5t. {/eq}

Which is a linear differential equation.

Where...

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