Find the solution of the differential equation $2′=^$ that satisfies the initial condition $(0)=2.$

Question

$$2y dy = xe^x dx$$

$$y = e^x (x-1) + C$$

Pluged in my initial condition $Y(0)=2. $ Got C = 3

Pluged C back into the equation and got a final answer below.

$$y = e^x (x-1) + 3$$

This answer is above is wrong! Please Help : )

Answer 1

Integrating both sides gives $$ \color{red}{y^2}=(x-1)e^x+C $$ We can determine $C$ with the initial condition: $2^2=(0-1)e^0+C$, so $C=5$.

Thus $y=\sqrt{(x-1)e^x+5}$. The solution is over $\mathbb{R}$.

Answer 2

Put $z=y^2$, the new unknown function of $x$. Then $z'=xe^x$, so $z=(x-1)e^x+C$. Hence $y^2=(x-1)e^x+C$ and $4=-1+C$ and $C=5$. We get $y^2=(x-1)e^x+5$ and $y=\sqrt{(x-1)e^x+5}$ by the initial condition ($y(0)=2>0$).