JEE Mains · Maths · STD 12 - 9. differential equations
Let the solution curve of the differential equation \(x d y=\left(\sqrt{x^{2}+y^{2}}+y\right) d x, x>0\), intersect the line \(x =1\) at \(y =0\) and the line \(x =2\) at \(y =\alpha\). Then the value of \(\alpha\) is.
- A \(\frac{1}{2}\)
- B \(\frac{3}{2}\)
- C \(-\frac{3}{2}\)
- D \(\frac{5}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
\(x d y=\left(\sqrt{x^{2}+y^{2}}+y\right) d x\) \(x d y-y d x=\sqrt{x^{2}+y^{2}} d x\) \(\frac{x d y-y d x}{x^{2}}=\sqrt{1+\frac{y^{2}}{x^{2}}} \cdot \frac{d x}{x}\) \(\frac{d(y / x)}{\sqrt{1+\left(\frac{y}{x}\right)^{2}}}=\frac{d x}{x}\)…
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