JEE Mains · Maths · STD 12 - 9. differential equations
If \(x = x ( y )\) is the solution of the differential equation \(y \frac{d x}{d y}=2 x+y^{3}(y+1) e^{y}, x(1)=0\); then \(x(e)\) is equal to
- A \(e^{3}\left(e^{e}-1\right)\)
- B \(e^{e}\left(e^{3}-1\right)\)
- C \(e ^{2}\left( e ^{ e }+1\right)\)
- D \(e ^{e}\left( e ^{2}-1\right)\)
Answer & Solution
Correct Answer
(A) \(e^{3}\left(e^{e}-1\right)\)
Step-by-step Solution
Detailed explanation
\(y \frac{d x}{d y}=2 x+y^{3}(y+1) e^{y}, x(1)=0\) \(\frac{d x}{d y}-\frac{2}{y} x=y^{2}(y+1) e^{y}\) I.f \(=e^{\int \frac{-2}{y} d y}=\frac{1}{y^{2}}\) \(x \cdot \frac{1}{y^{2}}=\int(y+1) e^{y} d y\) \(\frac{x}{y^{2}}=(y+1) e^{y}-e^{y}+c=y \cdot e^{y}+c\)…
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