# y=Pe^(ax) + Qe^(bx) Show that (d^2y)/dx^2-(a+b)dy/dx+aby=0?

Feb 28, 2018

See below

#### Explanation:

Let $y = P {e}^{a x} + Q {e}^{b x}$

Then

$a b y = a b \left(P {e}^{a x} + Q {e}^{b x}\right) = a b P {e}^{a x} + a b Q {e}^{b x}$

And $\frac{\mathrm{dy}}{\mathrm{dx}} = a P {e}^{a x} + b Q {e}^{b x}$

$\Rightarrow \left(a + b\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \left(a + b\right) \left(a P {e}^{a x} + b Q {e}^{b x}\right) = {a}^{2} P {e}^{a x} + a b Q {e}^{b x} + a b P {e}^{a x} + {b}^{2} Q {e}^{b x}$

And $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = {a}^{2} P {e}^{a x} + {b}^{2} Q {e}^{b x}$

So

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 - \left(a + b\right) \frac{\mathrm{dy}}{\mathrm{dx}} + a b y$
$= {a}^{2} P {e}^{a x} + {b}^{2} Q {e}^{b x} - \left({a}^{2} P {e}^{a x} + a b Q {e}^{b x} + a b P {e}^{a x} + {b}^{2} Q {e}^{b x}\right) + a b P {e}^{a x} + a b Q {e}^{b x} = 0 ,$ as required. $\square$

Feb 28, 2018

$\textcolor{red}{y = P {e}^{a x} + Q {e}^{b x} \Rightarrow {y}_{2} - \left(a + b\right) {y}_{1} + a b y = 0}$
Answer for above Question is given below.

#### Explanation:

We have,
$y = P {e}^{a x} + Q {e}^{b x}$
$\Rightarrow y = P {e}^{a x} + \frac{Q}{e} ^ \left(- b x\right)$
$\Rightarrow {e}^{- b x} y = P {e}^{a x - b x} + Q$
$\Rightarrow {e}^{- b x} y - P {e}^{a x - b x} = Q$, where Q is constant
Diff.w.r.t.x,we gate,
${e}^{- b x} {y}_{1} + y \left({e}^{- b x}\right) \left(- b\right) - P {e}^{a x - b x} \left(a - b\right) = 0$
Dividing both sides by, ${e}^{- b x}$
${y}_{1} - b y - P {e}^{a x} \left(a - b\right) = 0 \Rightarrow {y}_{1} - b y - \frac{P \left(a - b\right)}{e} ^ \left(- a x\right) = 0$
$\Rightarrow {e}^{- a x} {y}_{1} - b {e}^{- a x} y - P \left(a - b\right) = 0$,where P,a,b are constants.
Diff.w.r.t.x,we get,
$\Rightarrow {e}^{- a x} {y}_{2} + {y}_{1} {e}^{- a x} \left(- a\right) - b {e}^{- a x} {y}_{1} - b y {e}^{- a x} \left(- a\right) = 0$
Dividing both sides by, ${e}^{- a x}$
$\Rightarrow {y}_{2} - a {y}_{1} - b {y}_{1} - b y \left(- a\right) = 0$
$\Rightarrow {y}_{2} - \left(a + b\right) {y}_{1} + a b y = 0$.

Feb 28, 2018

Kindly refer to a Proof in the Explanation.

#### Explanation:

Prerequisite : The elimination of $p \mathmr{and} q$ from the eqns. :

$p x + q y + z = 0 , p l + q m + n = 0 , p u + q v + w = 0$ is,

$| \left(x , y , z\right) , \left(l , m , n\right) , \left(u , v , w\right) | = 0$.

We have, $P {e}^{a x} + q {e}^{b x} - y = 0. \ldots \ldots \ldots \left(1\right)$.

Diff.ing w.r.t. $x$, we get, $P \cdot a {e}^{a x} + Q \cdot b {e}^{b x} - y ' = 0. \ldots \ldots . \left(2\right)$.

Rediff.ing w.r.t. $x$, we get, $P \cdot {a}^{2} {e}^{a x} + Q \cdot {b}^{2} {e}^{b x} - y ' ' = 0. . . \left(3\right)$.

Eliminating $P \mathmr{and} Q$ from $\left(1\right) , \left(2\right) \mathmr{and} \left(3\right)$, we have,

$| \left({e}^{a x} , {e}^{b x} , - y\right) , \left(a {e}^{a x} , b {e}^{b x} , - y '\right) , \left({a}^{2} {e}^{a x} , {b}^{2} {e}^{b x} , - y ' '\right) | = 0$.

$\therefore {e}^{a x} {e}^{b x} | \left(1 , 1 , - y\right) , \left(a , b , - y '\right) , \left({a}^{2} , {b}^{2} , - y ' '\right) | = 0$.

Since ${e}^{a x} , {e}^{b x} \ne 0$, and applying ${C}_{2} - {C}_{1}$, we get,

$| \left(1 , 0 , - y\right) , \left(a , b - a , - y '\right) , \left({a}^{2} , {b}^{2} - {a}^{2} , - y ' '\right) | = 0$.

$\therefore \left(b - a\right) | \left(1 , 0 , - y\right) , \left(a , 1 , - y '\right) , \left({a}^{2} , b + a , - y ' '\right) | = 0$.

As $\left(b - a\right) \ne 0$, and applying ${R}_{2} - a \cdot {R}_{1} , \mathmr{and} {R}_{3} - {a}^{2} \cdot {R}_{1}$,

$| \left(1 , 0 , - y\right) , \left(0 , 1 , a y - y '\right) , \left(0 , b + a , {a}^{2} y - y ' '\right) | = 0$.

Finally, expanding by ${C}_{1}$, we have,

$1 \cdot | \left(1 , a y - y '\right) , \left(b + a , {a}^{2} y - y ' '\right) | - 0 + 0 = 0 , i . e . ,$

$1 \left({a}^{2} y - y ' '\right) - \left(b + a\right) \left(a y - y '\right) = 0 , \mathmr{and} ,$

${a}^{2} y - y ' ' - b a y - {a}^{2} y + \left(b + a\right) y ' = 0$.

$\Rightarrow y ' ' - \left(a + b\right) y ' + a b y = 0$ what is the same as to say that,

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 - \left(a + b\right) \frac{\mathrm{dy}}{\mathrm{dx}} + a b y = 0$.

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