JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution of the differential equation \(\frac{d y}{d x}=2 x(x+y)^3-x(x+y)-1, y(0)=1\). Then, \(\left(\frac{1}{\sqrt{2}}+\mathrm{y}\left(\frac{1}{\sqrt{2}}\right)\right)^2\) equals :
- A \(\frac{4}{4+\sqrt{\mathrm{e}}}\)
- B \(\frac{3}{3-\sqrt{\mathrm{e}}}\)
- C \(\frac{2}{1+\sqrt{\mathrm{e}}}\)
- D \(\frac{1}{2-\sqrt{\mathrm{e}}}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{2-\sqrt{\mathrm{e}}}\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=2 x(x+y)^5-x(x+y)-1\) \(x+y=t\) \(\frac{d t}{d x}-1=2 x t^3-x t-1\) \(\frac{d t}{2 t^3-t}=x d x\) \(\frac{t d t}{2 t^4-t^2}=x d x\) Let \(t^2=z\) \(\int \frac{d z}{2\left(2 z^2-z\right)}=\int x d x\) \(\int \frac{d z}{4 z\left(z-\frac{1}{2}\right)}=\int x d x\)…
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