MHT CET · Maths · Differential Equations
\(y=c^{2}+\frac{c}{x}\) is the solution of the differential equation
- A \(x^{4}\left(\frac{d y}{d x}\right)^{2}+x\left(\frac{d y}{d x}\right)-y=0\)
- B \(x^{4}\left(\frac{d y}{d x}\right)^{2}-x\left(\frac{d y}{d x}\right)-y=0\)
- C \(x^{4}\left(\frac{d y}{d x}\right)^{2}-x\left(\frac{d y}{d x}\right)+y=0\)
- D \(x^{4}\left(\frac{d y}{d x}\right)^{2}+x\left(\frac{d y}{d x}\right)+y=0\)
Answer & Solution
Correct Answer
(B) \(x^{4}\left(\frac{d y}{d x}\right)^{2}-x\left(\frac{d y}{d x}\right)-y=0\)
Step-by-step Solution
Detailed explanation
Given \(y=c^{2}+\frac{c}{x}\)
Differentiating w.r.t. \(x\)
\(
\frac{d y}{d x}=0-\frac{c}{x^{2}}=-\frac{c}{x^{2}} \Rightarrow c=\left(-x^{2}\right)\left(\frac{d y}{d x}\right)
\)
Substituting value of \(\mathrm{c}\) in given equation, We get
\(
\begin{aligned}
y &=\left[\left(-x^{2}\right)\left(\frac{d y}{d x}\right)\right]^{2}+\frac{\left(-x^{2}\right)\left(\frac{d y}{d x}\right)}{x} \\
&=x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x} \\
\therefore & x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}-y=0
\end{aligned}
\)
Differentiating w.r.t. \(x\)
\(
\frac{d y}{d x}=0-\frac{c}{x^{2}}=-\frac{c}{x^{2}} \Rightarrow c=\left(-x^{2}\right)\left(\frac{d y}{d x}\right)
\)
Substituting value of \(\mathrm{c}\) in given equation, We get
\(
\begin{aligned}
y &=\left[\left(-x^{2}\right)\left(\frac{d y}{d x}\right)\right]^{2}+\frac{\left(-x^{2}\right)\left(\frac{d y}{d x}\right)}{x} \\
&=x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x} \\
\therefore & x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}-y=0
\end{aligned}
\)
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