MHT CET · Maths · Differential Equations
\(y=m x+\frac{2}{m}\) is the general solution of
- A \(y\left(\frac{d y}{d x}\right)^{2}=x\left(\frac{d y}{d x}\right)+2\)
- B \(y=x \frac{d y}{d x}+2\)
- C \(y\left(\frac{d y}{d x}\right)=x\left(\frac{d y}{d x}\right)^{2}+2\)
- D \(y\left(\frac{d y}{d x}\right)=x+2\)
Answer & Solution
Correct Answer
(C) \(y\left(\frac{d y}{d x}\right)=x\left(\frac{d y}{d x}\right)^{2}+2\)
Step-by-step Solution
Detailed explanation
\(
\begin{array}{l}
\text { Given } \mathrm{y}=\mathrm{m} \mathrm{x}+\frac{2}{\mathrm{~m}}....(1) \\
\therefore \frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}=(\mathrm{m} \times 1)+0 \Rightarrow \mathrm{m}=\frac{\mathrm{dy}}{\mathrm{dx}}
\end{array}
\)
Putting value of \(m\) in equation (1), we get
\(
y=x \frac{d y}{d x}+\frac{2}{\left(\frac{d y}{d x}\right)} \Rightarrow y \frac{d y}{d x}=x\left(\frac{d y}{d x}\right)^{2}+2
\)
\begin{array}{l}
\text { Given } \mathrm{y}=\mathrm{m} \mathrm{x}+\frac{2}{\mathrm{~m}}....(1) \\
\therefore \frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}=(\mathrm{m} \times 1)+0 \Rightarrow \mathrm{m}=\frac{\mathrm{dy}}{\mathrm{dx}}
\end{array}
\)
Putting value of \(m\) in equation (1), we get
\(
y=x \frac{d y}{d x}+\frac{2}{\left(\frac{d y}{d x}\right)} \Rightarrow y \frac{d y}{d x}=x\left(\frac{d y}{d x}\right)^{2}+2
\)
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