Solve $$2yy' + 1 = y^2 + x, y(0) = 1$$
Before I even start this problem, how do you determine what to set v equal to?
I figured it out to be $$y^2 + x$$ Anyways, $$v = y^2 + x, v' = 2yy' + 1$$ $$v' = v$$ $$\frac{\,dv}{v} = \, dx$$ $$\ln|v| = x + C$$ $$v = De^x$$ where $$D = \pm1$$ Since $v = y^2,$ then $y = \pm\sqrt{De^x}$ $$y(0) = 1, \text{therefore}, D = 1$$
So does that mean the answer is $$y = \pm\sqrt{e^x}$$?
$$\begin{align}2yy'+1&=y^2+x\\v'&=v\\\dfrac{\mathrm dv}{\mathrm dx}&=v\\\int\dfrac{\mathrm dv}v&=\int \,\mathrm dx\\\ln|v|&=x+C\\v&=Ae^x\qquad\bigg(\because A=e^C\bigg)\\y^2+x&=Ae^x\\y^2&=Ae^x-x\end{align}$$
Using $y(0)=1$
$$\begin{align}1&=Ae^0-0\\A&=1\end{align}$$
Therefore the solution is $$y^2=e^x-x$$