Let y=y(x) be the solution of the differential equation (x y-5 x^2 √1+x^2) d x+(1+x^2) d y…

Question

Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be the solution of the differential equation $\left(x y-5 x^2 \sqrt{1+x^2}\right) d x+\left(1+x^2\right) d y=0, y(0)=0$. Then $y(\sqrt{3})$ is equal to

Accepted Answer

$$\begin{aligned}
& \left(1+x^2\right) \frac{d y}{d x}+x y=5 x^1 \sqrt{1+x^2} \
& \frac{d y}{d x}+\frac{x y}{1+x^2}=\frac{5 x^2}{\sqrt{1+x^2}} \
& \therefore \text { I.F. }=e^{\int \frac{x}{1+x^2} d x}=e^{\frac{\ln \left(1+x^2\right)}{2}}=\sqrt{1+x^2} \
& \therefore y \sqrt{1+x^2}=\int \frac{5 x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} d x \
& \therefore y \sqrt{1+x^2}=\int \frac{5 x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} d x \
& y \sqrt{1+x^2}=\frac{5 x^3}{3}+C \
& \because y(0)=0 \Rightarrow 0=0+C \Rightarrow C=0 \
& \therefore y=\frac{5 x^3}{3 \sqrt{1+x^2}} \
& y(\sqrt{3})=\frac{15 \sqrt{3}}{32}=\frac{5 \sqrt{3}}{2}
\end{aligned}$$

Solutions · 1