Trying to solve a differential equation

Question

I am trying to solve the following differential equation

$$y''=2yy'$$

and ended up with the answer: $$\frac{1}{C_1}\arctan\frac{y}{C_1}+C_2=x.$$

However, the answer should be,

$$\arctan\frac{y}{C_1}+C_2=x.$$

So I tried manipulating my solution in the following way,

$$\left(\frac{1}{C_1}\right)\left(\frac{y}{C_1}\right)=\tan(x-C_2)$$

$$\left(\frac{y}{C_1}\right)=\tan(x-C_2)$$ $$\arctan\frac{y}{C_1}+C_2=x.$$

Which ends up being the correct answer.

Did I break any law of math in the above steps? Or it can't be done that way? This is the only way I can think of on how my professor come up with such answer.

Answer 1

Your first answer is correct. The later manipulation you performed to get the answer you say your teacher came up with is wrong. $$ \left(\frac1{C_1}\right)\left(\frac{y}{C_1}\right)=\tan(x-C_2) $$ does not follow from $$ \frac1{C_1}\arctan\frac{y}{C_1}+C_2=x $$

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Additionally, there are two other answers depending on the sign of the constant of integration of the first integration.

Integrating $$ y''=2yy' $$ gives $$ y'=y^2+c_1 $$ Dividing and integrating yields $$ \int\frac{\mathrm{d}y}{y^2+c_1}=\int\mathrm{d}x $$ If $c_1\gt0$, $$ \frac1{\sqrt{c_1}}\tan^{-1}\left(\frac{y}{\sqrt{c_1}}\right)+c_2=x $$ which gives $$ \bbox[5px,border:2px solid #C0A000]{y=a\tan(ax+b)} $$ If $c_1=0$, $$ c_2-\frac1y=x $$ which gives $$ \bbox[5px,border:2px solid #C0A000]{y=\frac1{b-x}} $$ If $c_1\lt0$, $$ -\frac1{\sqrt{-c_1}}\tanh^{-1}\left(\frac{y}{\sqrt{-c_1}}\right)+c_2=x $$ which gives $$ \bbox[5px,border:2px solid #C0A000]{y=-a\tanh(ax+b)} $$

Answer 2

Make the susbtitution $$u(y)=y'(x)$$ So your differential equations becomes $$u'(y)\,u(y)=2y\,u(y)\Longleftrightarrow u(y)(2y-u'(y))=0.$$ Solving the above equation will result in: $$\begin{aligned}&(1)\,u(y)=0,\,\\&(2)\,u'(y)=2y.\end{aligned}$$

For $(2)$: $$u(y)=2\int y\,dy=2y^2+C_1.$$ Therefore, the solutions for $u$ are: $$u(y)=\{0,y^2+C_1\}.$$ Substituting back into the initial relation, you have $$(3)\,y'(x)=0, \qquad (4)\,y'(x)=y^2(x)+C_1$$ The solution to $(3)$ is simply a costant, while $(4)$: $$y'(x)=y^2(x)+C_1\Longleftrightarrow \frac{dy(x)}{y^2(x)+C_1}=dx$$ and by integrating both sides, you obtain: $$\operatorname{\LARGE\int}\frac{dy(x)}{\left(\frac{y(x)}{\sqrt{C_1}}\right)^2+1}=\operatorname{\LARGE\int} dx\implies\frac1{\sqrt{C_1}}\arctan\left(\frac{y(x)}{\sqrt{C_1}}\right)=x+C_2.$$ Solving for $y(x)$ gives $$y(x)=\sqrt{C_1}\tan\left(x\sqrt{C_1}+C_2\sqrt{C_1}\right).$$