MHT CET · Maths · Differential Equations
The differential equation of all family of lines \(y=m x+\frac{4}{m}\) obtained by eliminating the arbitrary constant \(\mathrm{m}\) is
- A \(y\left(\frac{d y}{d x}\right)=4\)
- B \(y\left(\frac{d y}{d x}\right)^2+y\left(\frac{d y}{d x}\right)+4=0\)
- C \(x\left(\frac{d y}{d x}\right)+4=0\)
- D \(x\left(\frac{d y}{d x}\right)^2-y\left(\frac{d y}{d x}\right)+4=0\)
Answer & Solution
Correct Answer
(D) \(x\left(\frac{d y}{d x}\right)^2-y\left(\frac{d y}{d x}\right)+4=0\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& y=m x+\frac{4}{m} \\
& \therefore \frac{d y}{d x}=m
\end{aligned}
\)
Substituting value of \(\mathrm{m}\) in equation (1), we get
\(
\begin{aligned}
& y=\left(\frac{d y}{d x}\right) x+\frac{4}{\left(\frac{d y}{d x}\right)} \\
& \therefore y\left(\frac{d y}{d x}\right)=\left(\frac{d y}{d x}\right)^2 x+4
\end{aligned}
\)
\begin{aligned}
& y=m x+\frac{4}{m} \\
& \therefore \frac{d y}{d x}=m
\end{aligned}
\)
Substituting value of \(\mathrm{m}\) in equation (1), we get
\(
\begin{aligned}
& y=\left(\frac{d y}{d x}\right) x+\frac{4}{\left(\frac{d y}{d x}\right)} \\
& \therefore y\left(\frac{d y}{d x}\right)=\left(\frac{d y}{d x}\right)^2 x+4
\end{aligned}
\)
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