Search any question & find its solution
Question:
Answered & Verified by Expert
Let $p \in I R$, then the differential equation of the family of curves $y=(\alpha+\beta x) e^{p x}$, where $\alpha, \beta$ are arbitrary constants, is
Options:
Solution:
1696 Upvotes
Verified Answer
The correct answer is:
$y^{\prime \prime}-2 p y^{\prime}+p^2 y=0$
Given, family of curve $y=(\alpha+\beta x) e^{p x}$
On differentiating Eq. (i) w.r.t. $x$, we get
$$
\begin{aligned}
\frac{d y}{d x} & =\beta \cdot e^{p x}+p e^{p x}(\alpha+\beta x) \\
\Rightarrow \quad y^{\prime} & =\beta \cdot e^{p x}+p y
\end{aligned}
$$
[by Eq. (i)]
On differentiating Eq. (ii) w.r.t. $x$, we get
$$
\begin{aligned}
\Rightarrow & \quad \frac{d^2 y}{d x^2} & =\beta \cdot p e^{p x}+p \frac{d y}{d x} \\
\Rightarrow & y^{\prime \prime} & =\beta \cdot p e^{p x}+p y^{\prime} \\
\Rightarrow & y^{\prime \prime} & =\frac{\left(y^{\prime}-P y\right)}{e^{p x}} \cdot p e^{p x}+p y^{\prime}
\end{aligned}
$$
[by Eq. (ii), $\left.\beta=\frac{\left(y^{\prime}-p y\right)}{e^{p x}}\right]$
$$
\begin{array}{rlrl}
& \Rightarrow & y^{\prime \prime} & =\left(y^{\prime}-p y\right) \cdot p+p y^{\prime} \\
\Rightarrow & y^{\prime \prime} & =p y^{\prime}-p^2 y+p y^{\prime} \\
& \Rightarrow & y^{\prime \prime}+p^2 y-2 p y^{\prime} & =0 \\
& \therefore & y^{\prime \prime}-2 p y^{\prime}+p^2 y & =0
\end{array}
$$
On differentiating Eq. (i) w.r.t. $x$, we get
$$
\begin{aligned}
\frac{d y}{d x} & =\beta \cdot e^{p x}+p e^{p x}(\alpha+\beta x) \\
\Rightarrow \quad y^{\prime} & =\beta \cdot e^{p x}+p y
\end{aligned}
$$
[by Eq. (i)]
On differentiating Eq. (ii) w.r.t. $x$, we get
$$
\begin{aligned}
\Rightarrow & \quad \frac{d^2 y}{d x^2} & =\beta \cdot p e^{p x}+p \frac{d y}{d x} \\
\Rightarrow & y^{\prime \prime} & =\beta \cdot p e^{p x}+p y^{\prime} \\
\Rightarrow & y^{\prime \prime} & =\frac{\left(y^{\prime}-P y\right)}{e^{p x}} \cdot p e^{p x}+p y^{\prime}
\end{aligned}
$$
[by Eq. (ii), $\left.\beta=\frac{\left(y^{\prime}-p y\right)}{e^{p x}}\right]$
$$
\begin{array}{rlrl}
& \Rightarrow & y^{\prime \prime} & =\left(y^{\prime}-p y\right) \cdot p+p y^{\prime} \\
\Rightarrow & y^{\prime \prime} & =p y^{\prime}-p^2 y+p y^{\prime} \\
& \Rightarrow & y^{\prime \prime}+p^2 y-2 p y^{\prime} & =0 \\
& \therefore & y^{\prime \prime}-2 p y^{\prime}+p^2 y & =0
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.