How to solve the differential equation $y''+2y(y')^2=\frac{2x+1}{x}y'$

Question

$$y''+2y(y')^2=\frac{2x+1}{x}y'$$ Not ordinary... I don't know that to do Please, help or give an advice

Answer 1

Start by dividing both sides by $y'(x)$, then you will obtain: $$\frac{y''(x)}{y'(x)}+2\cdot y(x)\cdot y'(x)=\frac{2x+1}{x}$$ We can now integrate both sides with respect to $x$: $$\color{green}{\int \frac{y''(x)}{y'(x)}~dx}+2\color{blue}{\int y(x)\cdot y'(x)~dx}=\int \left(2+\frac{1}{x}\right)~dx \tag{1}$$ For the green integral, substitute $u=y'(x)$. $$\int \frac{y''(x)}{y'(x)}~dx=\int \frac{du}{u}=\ln(u)+C=\ln(y'(x))+ C$$ For the blue integral, substitute $v=y(x)$. $$\int y(x)\cdot y'(x)~dx=\int v~dv=\frac{v^2}{2}+C=\frac{(y(x))^2}{2}+C$$ The right hand side integral is trivial. Therefore: $$\ln(y'(x))+(y(x))^2=2x+\ln{x}+C \iff y'(x)=k_1e^{2x+\ln{x}-(y(x))^2}$$ Separating gives: $$e^{y^2}\cdot \frac{dy}{dx}=k_1xe^{2x} \iff \int e^{y^2}~dy=\int k_1 xe^{2x}~dx \tag{2}$$

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The left hand side of $(2)$ cannot be integrated in terms of elementary functions. However, you can deduce from the definition of the imaginary error function $\operatorname*{erfi}(\cdot )$ that: $$\operatorname*{erfi}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{t^2}~dt \implies \int e^{y^2}~dy=\frac{\sqrt{\pi}}{2}\operatorname*{erfi}(y)+C$$ Hence, we have the following implicit solution after using integration by parts on the right integral: $$\bbox[5px,border:2px solid #C0A000]{\frac{\sqrt{\pi}}{2}\operatorname*{erfi}(y)=\frac{1}{4}k_1e^{2x}(2x-1)+k_2} \tag{3}$$

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We can obtain an explicit solution by using the inverse imaginary error function $\operatorname*{erfi}^{-1}(\cdot)$ $$\operatorname{erfi}(y)=\frac{k_1e^{2x}(2x-1)+4k_2}{2\sqrt{\pi}}$$ $$y(x)=\operatorname{erfi}^{-1}\left(\frac{k_1e^{2x}(2x-1)+4k_2}{2\sqrt{\pi}}\right)$$ $$\bbox[5px,border:2px solid #C0A000]{y(x)=\operatorname{erfi}^{-1}\left(Ae^{2x}\left(2x-1\right)+B\right)}$$ Where $A=\frac{k_1}{2\sqrt{\pi}}$ and $B=\frac{2k_2}{\sqrt{\pi}}$. This can be done since $k_1$ and $k_2$ are arbitrary constants.

Answer 2

Hint. Divide by $y'$. Then you get $$ \frac{y''}{y'}+2yy'=2+\frac{1}{x}. $$ Integrate and obtain $$ \ln y'+y^2=2x+\ln x+c $$ Hence $$ \mathrm{e}^{y^2}y'=c'x\,\mathrm{e}^{2x}. $$ Then integrate once again...