MHT CET · Maths · Differential Equations
The general solution of \(\left(x \frac{d y}{d x}-y\right) \sin \frac{y}{x}=x^3 e^x\) is
- A \(\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1)+\cos \frac{\mathrm{y}}{\mathrm{x}}+\mathrm{c}=0\)
- B \(\mathrm{xe}^{\mathrm{x}}+\cos \frac{\mathrm{y}}{\mathrm{x}}+\mathrm{c}=0\)
- C \(\mathrm{e}^{\mathrm{x}}(\mathrm{x}+1)+\cos \frac{\mathrm{y}}{\mathrm{x}}+\mathrm{c}=0\)
- D \(\mathrm{ex}^{\mathrm{x}}-\cos \frac{\mathrm{y}}{\mathrm{x}}+\mathrm{c}=0\)
Answer & Solution
Correct Answer
(A) \(\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1)+\cos \frac{\mathrm{y}}{\mathrm{x}}+\mathrm{c}=0\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \left(x \frac{d y}{d x}-y\right) \sin \frac{y}{x}=x^3 e^x \\
& \text { Put } \frac{y}{x}=t \Rightarrow \frac{x \frac{d y}{d x}-y}{x^2}=\frac{d t}{d x}
\end{aligned}
\)
Put \(\frac{y}{x}=t \Rightarrow \frac{x \frac{d y}{d x}-y}{x^2}=\frac{d t}{d x}\)
Hence eq. (i) becomes
\( \mathrm{x}^2\left(\frac{d t}{d x}\right) \sin t=x^3 e^x \Rightarrow\left(\frac{d t}{d x}\right) \sin t=x e^x \)
\( \therefore \int \sin t d t=\int x^x d x \)
\( \therefore-\cos t=x e^x-\int e^x d x=x e^4-e^x+c \)
\( \therefore-\cos \left(\frac{y}{x}\right)=e^x(x-1)+c \)
\begin{aligned}
& \left(x \frac{d y}{d x}-y\right) \sin \frac{y}{x}=x^3 e^x \\
& \text { Put } \frac{y}{x}=t \Rightarrow \frac{x \frac{d y}{d x}-y}{x^2}=\frac{d t}{d x}
\end{aligned}
\)
Put \(\frac{y}{x}=t \Rightarrow \frac{x \frac{d y}{d x}-y}{x^2}=\frac{d t}{d x}\)
Hence eq. (i) becomes
\( \mathrm{x}^2\left(\frac{d t}{d x}\right) \sin t=x^3 e^x \Rightarrow\left(\frac{d t}{d x}\right) \sin t=x e^x \)
\( \therefore \int \sin t d t=\int x^x d x \)
\( \therefore-\cos t=x e^x-\int e^x d x=x e^4-e^x+c \)
\( \therefore-\cos \left(\frac{y}{x}\right)=e^x(x-1)+c \)
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