MHT CET · Maths · Differential Equations
The particular solution of \(\frac{d y}{d x}=1+x+y^2+x y^2\), when \(y(0)=0\). is
- A \(y=\log \left(1+\frac{x^2}{2}\right)\)
- B \(y^3=\log \left(1+\frac{x^2}{2}\right)\)
- C \(y^2=\tan \left(1+\frac{x^2}{2}\right)\)
- D \(y=\tan \left(x+\frac{x^2}{2}\right)\)
Answer & Solution
Correct Answer
(D) \(y=\tan \left(x+\frac{x^2}{2}\right)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{d y}{d x}=1+x+y^2+x y^2=(1+x)\left(1+y^2\right) \\ & \Rightarrow \int \frac{d y}{1+y^2}=\int(1+x) d x \\ & \Rightarrow \tan ^{-1} y=x+\frac{x^2}{2}+c \\ & \because y(0)=0 \Rightarrow c=0 \\ & \Rightarrow \tan ^{-1} y=x+\frac{x^2}{2} \\ & \Rightarrow y=\tan \left(x+\frac{x^2}{2}\right)\end{aligned}\)
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