Question

Consider the first order separable equation

Answer 1

Answer with Step-by-step explanation:

We are given that first order separable equation

$yy'=x(1+y^2)$

y(0)=1

$y\frac{dy}{dx}=x(1+y^2)$

$\frac{y}{1+y^2}dy=xdx$

Taking integration on both sides

$\int\frac{y}{1+y^2}dy=\int xdx$

Suppose $1+y^2=u$

Differentiate w.r.t u

$2ydy=du$

$ydy=\frac{1}{2}du$

Substitute the values

$\frac{1}{2}\int \frac{du}{u}=\int xdx$

$\frac{1}{2}lnu+C=\frac{x^2}{2}$

By using the formula

$\int x^ndx=\frac{x^{n+1}}{n+1}$+C

$\int \frac{dx}{x}=ln x+C$

$ln(1+y^2)+2C=x^2$

$ln(1+y^2)+C'=x^2$

$x^2-ln(1+y^2)=C'$

Suppose $f(y)=-ln(1+y^2)$

Then, the solution

$x^2+f(y)=C'$

Substitute t=0

$0-ln(1+1)=C'$

$C'=-ln 2$

Substitute the values

$x^2=ln(1+y^2)-ln 2$

$x^2=ln\frac{1+y^2}{2}$

By using identity $ln x-ln y=ln\frac{x}{y}$

$\frac{1+y^2}{2}=e^{x^2}$

By using identity $lnx=y\implies x=e^y$

$1+y^2=2e^{x^2}$

$y^2=2e^{x^2}-1$

$y=\sqrt{2e^{x^2}-1}$