Third Order Differential Equation (Change of variable)

Question

I'm trying to find a suitable change of variable for

$yy'''+3y'y''=2e^x$. But no idea, I tried almost everything... e.g. $x=\log t$ but the problem is getting huge.

Any ideas?

Thanks

Answer 1

$$yy'''+3y'y''=2e^x$$ $$yy'''+y'y''+2y'y''=2e^x$$ $$(yy'')'+2y'y''=2e^x$$ $$(yy'')'+(y'^2)'=2e^x$$ Integrate $$yy''+y'^2=2e^x+K_1$$ $$(y'y)'=2e^x+K_1$$ Integrate again $$y'y=2e^x+K_1x+K_2$$ $$(y^2)'=4e^x+C_1x+C_2$$ Integrate again.... $$y^2=4e^x+R_1x^2+R_2x+R_3$$ $$\boxed {y(x)=\pm \sqrt {4e^x+R_1x^2+R_2x+R_3}}$$

Answer 2

First of all, notice that:

\begin{align*} &yy^{\prime\prime\prime} + 3y^{\prime}y^{\prime\prime} = 2e^{x} \Leftrightarrow \int yy^{\prime\prime\prime}\mathrm{d}x + 3\int y^{\prime}y^{\prime\prime}\mathrm{d}x = 2\int e^{x}\mathrm{d}x \Leftrightarrow\\ & yy^{\prime\prime} - \int y^{\prime}y^{\prime\prime}\mathrm{d}x + 3\int y^{\prime}y^{\prime\prime}\mathrm{d}x = 2e^{x} \Leftrightarrow yy^{\prime\prime} + (y^{\prime})^{2} = 2e^{x} + k \end{align*} If we repeat the process, we get: \begin{align*} &\int yy^{\prime\prime}\mathrm{d}x + \int y^{\prime}y^{\prime}\mathrm{d}x = 2\int e^{x}\mathrm{d}x + \int k\mathrm{d}x \Leftrightarrow\\ & yy^{\prime} - \int y^{\prime}y^{\prime}\mathrm{d}x + \int y^{\prime}y^{\prime}\mathrm{d}x = 2e^{x} + kx \Leftrightarrow yy^{\prime} = 2e^{x} + kx + c \Leftrightarrow\\ &\int yy^{\prime}\mathrm{d}x = 2e^{x} + \frac{kx^{2}}{2} + cx + d \Leftrightarrow (y)^{2} = 4e^{x} + kx^{2} + 2cx + 2d \Leftrightarrow\\ & y(x) = \pm\sqrt{4e^{x} + kx^{2} + 2cx + 2d} \end{align*}