MCQ 11 Mark
Assertion (A) : If $\frac{d y}{d x}+x y=x^3 y^3, x>0, y \geq 0$ and $y(0)=1$, then $y(1)=\frac{1}{\sqrt{2}}$.
Reason (R) : The differential equation is linear with integrating factor $e^x$.
Reason (R) : The differential equation is linear with integrating factor $e^x$.
- ABoth (A) and (R) are true and (R) is the correct explanation of (A).
- BBoth (A) and (R) are true but (R) is not the correct explanation of (A).
- C(A) is true but (R) is false.
- D(A) is false but (R) is true.
Answer
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\begin{array}{l}
\text { (c) : } \frac{1}{y^3} \frac{d y}{d x}+\frac{x}{y^2}=x^3 \\
\text { Put } \frac{1}{y^2}=z \Rightarrow-\frac{2}{y^3} d y=d z \\
\therefore \quad \frac{d z}{d x}-2 x z=-2 x^3,
\end{array}
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which is a linear differential equation with I.F. $=e^{-x^2}$
$
\begin{array}{l}
\therefore \quad \text { Solution, } z e^{-x^2}=-\int e^{-x^2} 2 x^3 d x+C \\
\Rightarrow z e^{-x^2}=\left(x^2+1\right) e^{-x^2}+C \\
\Rightarrow \quad z=x^2+1+C e^{x^2} \\
\therefore \quad \frac{1}{y^2}=x^2+1+C e^{x^2} \\
\because \quad y(0)=1 \Rightarrow C=0 \\
\therefore \quad y^2=\frac{1}{x^2+1} \Rightarrow y=\frac{1}{\sqrt{x^2+1}} \\
\Rightarrow \quad y(1)=\frac{1}{\sqrt{2}}
\end{array}
$
\begin{array}{l}
\text { (c) : } \frac{1}{y^3} \frac{d y}{d x}+\frac{x}{y^2}=x^3 \\
\text { Put } \frac{1}{y^2}=z \Rightarrow-\frac{2}{y^3} d y=d z \\
\therefore \quad \frac{d z}{d x}-2 x z=-2 x^3,
\end{array}
$
which is a linear differential equation with I.F. $=e^{-x^2}$
$
\begin{array}{l}
\therefore \quad \text { Solution, } z e^{-x^2}=-\int e^{-x^2} 2 x^3 d x+C \\
\Rightarrow z e^{-x^2}=\left(x^2+1\right) e^{-x^2}+C \\
\Rightarrow \quad z=x^2+1+C e^{x^2} \\
\therefore \quad \frac{1}{y^2}=x^2+1+C e^{x^2} \\
\because \quad y(0)=1 \Rightarrow C=0 \\
\therefore \quad y^2=\frac{1}{x^2+1} \Rightarrow y=\frac{1}{\sqrt{x^2+1}} \\
\Rightarrow \quad y(1)=\frac{1}{\sqrt{2}}
\end{array}
$