Question

Solve differential equation(y2+1)dx=ysec^2(x)dy, y(0)=0

First order differential equations
ANSWERED
asked 2021-02-09

Solve differential equation \((y2+1)dx=y \sec^2(x)dy\), \(y(0)=0\)

Answers (1)

2021-02-10

\((y^2+1)=(y \sec^2x)dy/dx\)
\((y^2+1)=(y \sec^2x)y'\)
\(y/(-y^2-1)y'= (-1)/\sec^2x\)
From the above, it is observed that the first order differential is in the form of \(N(x)y'=M(x)\), where \(N(x)= y/(-y^2-1)\) and \(M(x) = (-1)/\sec^2x\)
Integrate on both sides of \(y/(-y^2-1)dy= (-1)/\sec^2x dx\)
\(\Rightarrow \int y/(-y^2-1) dy= \int (-1)/\sec^2x dx\)
\(\Rightarrow -\int y/(-y^2-1) dy= -\int 1/\sec^2x dx\)
\(\Rightarrow -1/2 \ln(-y^2-1)= /1/2(x+1/2\sin(2x))+c\)
\(-1/2 \ln(0-1)= -1/2(0+1/2\sin(0))+c\) \([:' y(0)=0]\)
\(c=0\) Substitute \(c=0 \in -1/2 \ln(-y^2-1)\pm1/2(x+1/2\sin(2x))+c\) and solve for y.
\(-1/2 \ln(-y^2-1)= -1/2(x+1/2\sin(2x))\)
\(\ln(-y^2-1)= x+1/2\sin(2x)\)
\(-y^2-1= e^{x+1/2\sin(2x)}\)
\(-y^2= e^{x+1/2\sin(2x)}+1\)
\(y^2= e^{x+1/2\sin(2x)}-1\)
\(y= \pm \sqrt{e^{x+1/2\sin(2x)}-1}\)

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