Question

Solve differential equation(cos^2y)/(4x+2)dy= ((cosy+siny)^2)/(sqrt(x^2+x+3))dx

First order differential equations
ANSWERED
asked 2021-03-07

Solve differential equation \(\frac{\cos^2y}{4x+2}dy= \frac{(\cos y+\sin y)^2}{\sqrt{x^2+x+3}}dx\)

Expert Answers (1)

2021-03-08

\(\frac{\cos^2y}{4x+2}dy=\frac{\cos^2y+\sin^2y+2\sin y \cos y}{\sqrt{x^2+x+3}}dx\)
\(\frac{\cos^2y}{4x+2}dy=\frac{1+\sin2y}{\sqrt{x^2+x+3}}dx\)
\(\frac{\cos2y}{1+\sin2y}dy= \frac{4x+2}{\sqrt{x^2+x+3}}dx\)
Integrating on both sides
\(\int \frac{\cos2y}{1+\sin2y}dy= \int \frac{4x+2}{\sqrt{x2+x+3}}dx\)
\(\int \frac{4x+2}{\sqrt{x2+x+3}}dx\)
\(Let\ x^2+x+3=u^2\)
\(2x+1dx= 2u du\)
\(4x+2dx= 4u du\)
\(\int \frac{4x+2}{\sqrt{x2+x+3}}dx= \frac{4u\ du}{u} du\)
\(= 4du= 4u+c\)
\(= 4 \sqrt{x^2+x+3}+c\)
\(\int \frac{\cos2y}{1+\sin2y}y\)
\(Let\ t=1+\sin2y\)
Differentiating
\(dt= 2\cos2y\)
\(\int \frac{\cos2y}{1+\sin2y}dy= \int \frac{\frac{1}{2}dt}{t}\)
\(= \frac{1}{2} \int \frac{1}{t} dt\) \(= \frac{1}{2} \ln t+K\)
\(\frac{1}{2} \ln(1+\sin2y)+k\)
\(\int \frac{\cos2y}{1+\sin2y}dy= \int\frac{4x+2}{\sqrt{x2+x+3}}dx\)
\(4 \sqrt{x2+x+3}= \frac{1}{2} \ln(1+\sin2y)+c\)

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