Let $y=y(x)$ be the solution of the differential equation $$\left(1-x^2\right) \mathrm{d} y=\left[x y+\left(x^3+2\right) \sqrt{3\left(1-x^2\right)}\right] \mathrm{d} x, -1< x<1, y(0)=0$$. If $y\left(\frac{1}{2}\right)=\frac{\mathrm{m}}{\mathrm{n}}, \mathrm{m}$ and $\mathrm{n}$ are co-prime numbers, then $\mathrm{m}+\mathrm{n}$ is equal to __________.
Answer (integer)
97
Solution
<p>$$\begin{aligned}
& \frac{d y}{d x}-\frac{x y}{1-x^2}=\frac{\left(x^3+2\right) \sqrt{3\left(1-x^2\right)}}{1-x^2} \\
& \mathrm{IF}=e^{-\int \frac{x}{1-x^2} d x}=e^{+\frac{1}{2} \ln \left(1-x^2\right)}=\sqrt{1-x^2} \\
& y \sqrt{1-x^2}=\sqrt{3} \int\left(x^3+2\right) d x \\
& y \sqrt{1-x^2}=\sqrt{3}\left(\frac{x^4}{4}+2 x\right)+c \\
& \Rightarrow y(0)=0 \quad \therefore c=0\\
& y\left(\frac{1}{2}\right)=\frac{65}{32}=\frac{m}{n} \\
& m+n=97
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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