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
Solution
<p>$$\begin{aligned}
& \left(1+x^2\right) \frac{d y}{d x}+x y=5 x^1 \sqrt{1+x^2} \\
& \frac{d y}{d x}+\frac{x y}{1+x^2}=\frac{5 x^2}{\sqrt{1+x^2}} \\
& \therefore \text { I.F. }=e^{\int \frac{x}{1+x^2} d x}=e^{\frac{\ln \left(1+x^2\right)}{2}}=\sqrt{1+x^2} \\
& \therefore y \sqrt{1+x^2}=\int \frac{5 x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} d x \\
& \therefore y \sqrt{1+x^2}=\int \frac{5 x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} d x \\
& y \sqrt{1+x^2}=\frac{5 x^3}{3}+C \\
& \because y(0)=0 \Rightarrow 0=0+C \Rightarrow C=0 \\
& \therefore y=\frac{5 x^3}{3 \sqrt{1+x^2}} \\
& y(\sqrt{3})=\frac{15 \sqrt{3}}{32}=\frac{5 \sqrt{3}}{2}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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