Question

Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b:$y=ae^{3x}+be^{-2x}$

Answer 1

$y=ae^{3x}+be^{-2x}$...(1)
Differentiating both sides with respect to x, we get:

$y'=3ae^{3x}-2be^{-2x}$...(2)

Again, differentiating both sides with respect to x, we get:

$y''=9ae^{3x}+4be^{-2x}...(3)$

Now, multiplying equation(1)with equation(2)and adding to equation(2), we get:

$(2ae^{3x}+2be^{-2x})+(3ae^{3x}-2be^{-2x})=2y+y'$

$\Rightarrow 5ae^{3x}=2y+y'$
$\Rightarrow ae^{3x}=\frac{2y+y'}{5}$
Now, multiplying equation(1)with equation(3)and subtracting equation(2)from it, we get:

(3ae3x+3be-2x)-(3ae3x-2be-2x)=3y-y'
$\Rightarrow$ 5be-2x=3y-y'
$\Rightarrow -2x=\frac{3y-y'}{5}$

Substituting the values of ae3x and be-2x in equation(3),we get:

$y''=9.\frac{(2y+y')}{5}+4\frac{(3y-y')}{5}$

$\Rightarrow y'' =\frac{18y+9y'}{5}+\frac{12y-4y'}{5}$

$\Rightarrow y''=\frac{30y+5y}{5}$

$\Rightarrow$ y''=6y+y'

$\Rightarrow$ y''-y'-6y=0

This is the required differential equation of the given curve.