JEE Mains · Maths · STD 12 - 9. differential equations
Let us consider a curve, \(\mathrm{y}=\mathrm{f}(\mathrm{x})\) passing through the point \((-2,2)\) and the slope of the tangent to the curve at any point \((x, f(x))\) is given by \(f(x)+x f^{\prime}(x)=x^{2}\) Then :
- A \(x^{2}+2 x\, f(x)-12=0\)
- B \(x^{3}+x \,f(x)+12=0\)
- C \(x^{3}-3 x\, f(x)-4=0\)
- D \(x^{2}+2 x\, f(x)+4=0\)
Answer & Solution
Correct Answer
(C) \(x^{3}-3 x\, f(x)-4=0\)
Step-by-step Solution
Detailed explanation
\(y+\frac{x d y}{d x}=x^{2}(\text { given })\) \(\Rightarrow \frac{d y}{d x}+\frac{y}{x}=x\) \(\text { If }=e^{\int \frac{1}{x} d x}=x\) Solution of \(\mathrm{DE}\) \(\Rightarrow y \cdot x=\int x \cdot x d x\) \(\Rightarrow x y=\frac{x^{3}}{3}+\frac{c}{3}\) Passes through…
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