Let $\mathrm{a}$ and $\mathrm{b}$ be arbitrary constants and $\mathrm{C}$ be a fixed constant. If $y=a e^{2…
Let $\mathrm{a}$ and $\mathrm{b}$ be arbitrary constants and $\mathrm{C}$ be a fixed constant. If $y=a e^{2 x}+b x e^{2 x}+C$ is the general solution of a differential equation, then the order of that differential equation is
$1$
$2$
$3$
$4$
Solution
Given $y=a e^{2 x}+b x e^{2 x}+c$ ...(i)
$\Rightarrow \quad y_1=2 a e^{2 x}+b\left[2 x e^{2 x}+e^{2 x}\right]$ ...(ii)
$\Rightarrow \quad y_2=4 a e^{2 x}+b\left[4 x e^{2 x}+4 e^{2 x}\right]$ ...(iii)
Now eliminating arbitrary constant $a, b$ from eqs (i), (ii) \& (iii), we get the require differential equations.
$\left|\begin{array}{ccc}y-c & -e^{2 x} & -x e^{2 x} \\ y_1 & -2 e^{2 x} & -(2 x+1) e^{2 x} \\ y_2 & -4 e^{2 x} & -(4 x+4) e^{2 x}\end{array}\right|=0$ ...(iv)
Since, here $\mathrm{c}$ is not a arbitrary constant, ' $\mathrm{c}$ ' is a fixed constant. Hence solving eq. (iv) will give a differential equation of second degree.
$
\begin{aligned}
& \Rightarrow\left|\begin{array}{ccc}
y-c & 1 & x \\
y_1 & 2 & (2 x+1) \\
y_2 & 4 & (4 x+4)
\end{array}\right|=0\left\{\because \text { dividing by }\left(-e^{2 x}\right)^2\right. \\
& \Rightarrow y_2+(8 x+4) y_1+4(y-c)=0 \\
& \Rightarrow \quad 2^{\text {nd }} \text { order diff. Eq. }
\end{aligned}
$