JEE Mains · Maths · STD 12 - 9. differential equations
Let the solution curve \(y = y ( x )\) of the differential equation \(\quad \frac{d y}{d x}-\frac{3 x^5 \tan ^{-1}\left(x^3\right)}{\left(1+x^6\right)^{\frac{3}{2}}} y=2 x\) \(\exp \frac{x^3-\tan ^{-1} x^3}{\sqrt{(1+x)^6}}\) pass through the origin. Then \(y (1)\) is equal to:
- A \(\exp \left(\frac{4-\pi}{4 \sqrt{2}}\right)\)
- B \(\exp \left(\frac{\pi-4}{4 \sqrt{2}}\right)\)
- C \(\exp \left(\frac{1-\pi}{4 \sqrt{2}}\right)\)
- D \(\exp \left(\frac{4+\pi}{4 \sqrt{2}}\right)\)
Answer & Solution
Correct Answer
(A) \(\exp \left(\frac{4-\pi}{4 \sqrt{2}}\right)\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}+\left(\frac{-3 x^5 \tan ^{-1} x^3}{\left(1+x^6\right)^{3 / 2}}\right) y=2 e^{\left\{\frac{x-\tan x}{\sqrt{1+x^6}}\right\}}\) \(\text { I.F. }=e^{\int \frac{-3 x^5 \tan ^{-1} x^3}{\left(1+x^6\right)^{3 / 2}} d x} \)…
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