JEE Mains · Maths · STD 12 - 9. differential equations
If \(y=y(x)\) is the solution curve of the differential equation \(\left(x^2-4\right) d y-\left(y^2-3 y\right) d x=0\), \(x>2, y(4)=\frac{3}{2}\) and the slope of the curve is never zero, then the value of \(y(10)\) equals :
- A \(\frac{3}{1+(8)^{1 / 4}}\)
- B \(\frac{3}{1+2 \sqrt{2}}\)
- C \(\frac{3}{1-2 \sqrt{2}}\)
- D \(\frac{3}{1-(8)^{1 / 4}}\)
Answer & Solution
Correct Answer
(A) \(\frac{3}{1+(8)^{1 / 4}}\)
Step-by-step Solution
Detailed explanation
\( \left(x^2-4\right) d y-\left(y^2-3 y\right) d x=0 \) \( \Rightarrow \int \frac{d y}{y^2-3 y}=\int \frac{d x}{x^2-4} \) \( \Rightarrow \frac{1}{3} \int \frac{y-(y-3)}{y(y-3)} d y=\int \frac{d x}{x^2-4} \)…
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