Question

Write a differential equation in wich $y^2=4(t+1)$ is a solutiion.

Answer 1

Solution

we have $y^2=4(t+1)\quad (1)$

$y^2=4t+4\implies 2y\frac{dy}{dt}=4\implies y\frac{dy}{dt}=2$

$\implies y'=\frac{2}{y}\iff \frac{dy}{dt}=\frac{2}{y}\iff y=dy=2dt$

$\iff \int ydy=2\int dt\implies \frac{1}{2}y^2+c_1=2t+c_2$

$\iff y^2=4t+2(c_2-c_1)\implies y^2=4t+c\quad (2)$

$\implies\begin{cases} y^2(0)=c\quad (2) & \quad \\ y^2(0)=4\quad (1) & \quad \end{cases}\implies c=4$

Then $y'=4t+4$ is a soltion of  $y^2-\frac{2}{y}=0$

Therefore, the differential equation is $y'-\frac{2}{y}=0$