Question

Form a differential equation representing the given family of curves by eliminating arbitrary constants $a$ and $b$
$y=e^{2x}(a+bx)$

Answer 1

let

$\frac{dy}{dx}=y^{\prime }\text{and}\frac{d^{2}y}{d{x}^2}=y^{\prime }$

Let the curve

$y=e^{2x}(a+bx)$

Differentiating both sides w.r.t. x
we get,

$y^{\prime }=\frac{d}{dx}\left[e^{2x}\right[a+bx\left]\right]$

Product rule $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$

$y^{\prime }=\frac{d\left[e^{2x}\right]}{dx}.[a+bx]+e^{2x}\frac{d[a+bx]}{dx}$

$y^{\prime }=2e^{2x}[a+bx]+e^{2x}.b$

$y^{\prime }=e^{2x}[2a+2bx+b]$

Again, differentiating both sides w.r.t. x
we get,
Product rule, $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$

$y"=\frac{d\left(e^{2x}\right)}{dx}[2a+2bx+b]+e^{2x}\frac{d[2a+2bx+b]}{dx}$

$y"=2e^{2x}[2a+2bx+b]+e^{2x}\times 2b$

Putting $y^{\prime }=e^{2x}[2a+2bx+b]$

$y"=2y^{\prime }+2e^{2x}b$

$y"-2y^{\prime }=2e^{2x}b$ (i)

Also,

$y^{\prime }-2y$

$=e^{2x}[2a+2bx+b]-2e^{2x}(a+bx)y^{\prime }-2y$

$=2a{e}^{2x}+2bx{e}^{2x}+e^{2x}b-2a{e}^{2x}-2bx{e}^{2x}$

$y^{\prime }-2y=(2a{e}^{2x}-2a{e}^{2x})+(2bx{e}^{2x}-2bx{e}^{2x})+e^{2x}b$

$y^{\prime }-2y=0+0+e^{2x}b$

$y^{\prime }-2y=e^{2x}b$

Now (𝑖) divided by (𝑖𝑖) then we get,

$\frac{y"-2y}{y^{\prime }-2y}=\frac{2e_b^{2x}}{e^{2x}b}$

$\frac{y"-2y^{\prime }}{y^{\prime }-2y}=2$

$y"-2y^{\prime }=2(y^{\prime }-2y)$

$y"-2y^{\prime }=2y^{\prime }-4y$

$y^{\prime }-2y^{\prime }-2y^{\prime }+4y=0$

$y^{\prime }-4y^{\prime }+4y=0$

Final Answer:
Hence, the required differential equation is

$y"-4y^{\prime }+4y=0$