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Differential Equations question

2024 · 30 Jan · Shift 1 · Q52
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  5. /2024 · 30 Jan · Shift 1 · Q52

Differential Equations question

2024 · 30 Jan · Shift 1 · Q52

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (1−x2)dy=[xy+(x3+2)3(1−x2)]dx,−1<x<1,y(0)=0\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\lt x\lt 1, y(0)=0(1−x2)dy=[xy+(x3+2)3(1−x2)​]dx,−1<x<1,y(0)=0. If y(12)=mn,my\left(\frac{1}{2}\right)=\frac{\mathrm{m}}{\mathrm{n}}, \mathrm{m}y(21​)=nm​,m and n\mathrm{n}n are co-prime numbers, then m+n\mathrm{m}+\mathrm{n}m+n is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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