Answer with Step-by-step explanation:
We are given that first order separable equation
[tex]yy'=x(1+y^2)[/tex]
y(0)=1
[tex]y\frac{dy}{dx}=x(1+y^2)[/tex]
[tex]\frac{y}{1+y^2}dy=xdx[/tex]
Taking integration on both sides
[tex]\int\frac{y}{1+y^2}dy=\int xdx[/tex]
Suppose [tex]1+y^2=u[/tex]
Differentiate w.r.t u
[tex]2ydy=du[/tex]
[tex]ydy=\frac{1}{2}du[/tex]
Substitute the values
[tex]\frac{1}{2}\int \frac{du}{u}=\int xdx[/tex]
[tex]\frac{1}{2}lnu+C=\frac{x^2}{2}[/tex]
By using the formula
[tex]\int x^ndx=\frac{x^{n+1}}{n+1}[/tex]+C
[tex]\int \frac{dx}{x}=ln x+C[/tex]
[tex]ln(1+y^2)+2C=x^2[/tex]
[tex]ln(1+y^2)+C'=x^2[/tex]
[tex]x^2-ln(1+y^2)=C'[/tex]
Suppose [tex]f(y)=-ln(1+y^2)[/tex]
Then, the solution
[tex]x^2+f(y)=C'[/tex]
Substitute t=0
[tex]0-ln(1+1)=C'[/tex]
[tex]C'=-ln 2[/tex]
Substitute the values
[tex]x^2=ln(1+y^2)-ln 2[/tex]
[tex]x^2=ln\frac{1+y^2}{2}[/tex]
By using identity [tex]ln x-ln y=ln\frac{x}{y}[/tex]
[tex]\frac{1+y^2}{2}=e^{x^2}[/tex]
By using identity [tex]lnx=y\implies x=e^y[/tex]
[tex]1+y^2=2e^{x^2}[/tex]
[tex]y^2=2e^{x^2}-1[/tex]
[tex]y=\sqrt{2e^{x^2}-1}[/tex]
