JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution of the differential equation \(x \tan \left(\frac{y}{x}\right) d y=\left(y \tan \left(\frac{y}{x}\right)-x\right) d x,-1 \leq x \leq 1, y\left(\frac{1}{2}\right)=\frac{\pi}{6}\) Then the area of the region bounded by the curves \(x=0, x=\frac{1}{\sqrt{2}}\) and \(y=y(x)\) in the upper half plane is :
- A \(\frac{1}{12}(\pi-3)\)
- B \(\frac{1}{6}(\pi-1)\)
- C \(\frac{1}{8}(\pi-1)\)
- D \(\frac{1}{4}(\pi-2)\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{8}(\pi-1)\)
Step-by-step Solution
Detailed explanation
We have \(\tan \left(\frac{y}{x}\right)(x d y-y d x)=-x d x\) \(\tan \left(\frac{y}{x}\right)\left(\frac{x d y-y d x}{x^{2}}\right)=-\frac{x}{x^{2}} d x\) \(\int \tan \left(\frac{y}{x}\right)\left(d\left(\frac{y}{x}\right)\right)=\int-\frac{1}{x} d x\)…
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