Obtain the differential equation by eliminating arbitrary constan…

Question

Obtain the differential equation by eliminating arbitrary constants from the following equations : $y=c_1 e^{3 x}+c_2 e^{2 x}$

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Answer

$
y=c_1 e^{3 x}+c_2 e^{2 x}
$
Differentiating twice w.r.t. $x$, we get
$
\begin{aligned}
& \frac{d y}{d x}=3 c_1 e^{3 x}+2 c_2 e^{2 x} \\
& \frac{d^2 y}{d x^2}=9 c_1 e^{3 x}+4 c_2 e^{2 x}
\end{aligned}
$
These three equations in $c_1 e^{3 x}$ and $c_2 e^{2 x}$ are consistent. $\therefore$ determinant of their consistency condition is zero.
$
\begin{aligned}
& \therefore\left|\begin{array}{ccc}
y & 1 & 1 \\
\frac{d y}{d x} & 3 & 2 \\
\frac{d^2 y}{d x^2} & 9 & 4
\end{array}\right|=0 \\
& \therefore y(12-18)-1\left(4 \frac{d y}{d x}-2 \frac{d^2 y}{d x^2}\right)+1\left(9 \frac{d y}{d x}-3 \frac{d^2 y}{d x^2}\right)=0 \\
& \therefore-6 y-4 \frac{d y}{d x}+2 \frac{d^2 y}{d x^2}+9 \frac{d y}{d x}-3 \frac{d^2 y}{d x^2}=0 \\
& \therefore \frac{d^2 y}{d x^2}-5 \frac{d y}{d x}+6 y=0 \\
\end{aligned}
$
This is the required D.E.
Alternative Method :
$
y=c_1 e^{3 x}+c_2 e^{2 x}
$
Dividing both sides by $e^{2 x}$, we get
$
e^{-2 x} y=c_1 e^x+c_2
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
& e^{-2 x} \frac{d y}{d x}+y \cdot e^{-2 x}(-2)=c_1 e^x+0 \\
& \therefore e^{-2 x}\left(\frac{d y}{d x}-2 y\right)=c_1 e^x
\end{aligned}
$
Dividing both sides by $e^x$, we get
$
e^{-3 x}\left(\frac{d y}{d x}-2 y\right)=c_1
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
& e^{-3 x}\left(\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}\right)+\left(\frac{d y}{d x}-2 y\right) \cdot e^{-3 x}(-3)=0 \\
& \therefore e^{-3 x}\left(\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}-3 \frac{d y}{d x}+6 y\right)=0 \\
& \therefore \frac{d^2 y}{d x^2}-5 \frac{d y}{d x}+6 y=0
\end{aligned}
$
This is the required D.E.