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QJEEM25Ma552b: JEE Main 2025 Mathematics - Class 12 JEEM25Ma552b Solution: Let be the solution of the differential equation . Then is…

Let y=y(x)\mathrm{y}=\mathrm{y}(\mathrm{x}) be the solution of the differential equation (xy−5x21+x2)dx+(1+x2)dy=0,y(0)=0\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(3)y(\sqrt{3}) is equal to

Multiple Choice Options

  • A 532\frac{5 \sqrt{3}}{2}
  • B 152\sqrt{\frac{15}{2}}
  • C 143\sqrt{\frac{14}{3}}
  • D 222 \sqrt{2}

Step-by-Step Verified Solution

amp;(1+x2)dydx+xy=5x11+x2amp;dydx+xy1+x2=5x21+x2amp;∴ I.F. =e∫x1+x2dx=eln⁡(1+x2)2=1+x2amp;∴y1+x2=∫5x21+x2⋅1+x2dxamp;∴y1+x2=∫5x21+x2⋅1+x2dxamp;y1+x2=5x33+Camp;∵y(0)=0⇒0=0+C⇒C=0amp;∴y=5x331+x2amp;y(3)=15332=532\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}

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