# If a particular integral of the differential equation (D^2+2D-1)y=e^(ax) is (-4/7)e^(ax) then the value of a is ?

Apr 16, 2018

$a = - \frac{3}{2} , - \frac{1}{2}$ 

We cannot eliminate a solution, thus we are left with two possibilities

$y = A {e}^{\left(- 1 - \sqrt{2}\right) x} + B {e}^{\left(- 1 + \sqrt{2}\right) x} - \frac{4}{7} {e}^{- \frac{3}{2} x}$

or

$y = A {e}^{\left(- 1 - \sqrt{2}\right) x} + B {e}^{\left(- 1 + \sqrt{2}\right) x} - \frac{4}{7} {e}^{- \frac{1}{2} x}$

#### Explanation:

We have:

$\left({D}^{2} + 2 D - 1\right) y = {e}^{a x}$ with a PI, $- \frac{4}{7} {e}^{a x}$

Or, In standard form:

$y ' ' + 2 y ' - y = {e}^{a x}$

This is a second order linear non-Homogeneous Differentiation Equation with constant coefficients. The standard approach is to find a solution, ${y}_{c}$ of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, ${y}_{p}$ of the non-homogeneous equation.

Complementary Function

The homogeneous equation associated with [A] is

$y ' ' + 2 y ' - y = 0$

And it's associated Auxiliary equation is:

${m}^{2} + 2 m - 1 = 0 \implies {\left(m + 1\right)}^{2} - 1 - 1 = 0$

Which has two real and distinct solution $m = - 1 \pm \sqrt{2}$

Thus the solution of the homogeneous equation is:

${y}_{c} = A {e}^{\left(- 1 - \sqrt{2}\right) x} + B {e}^{\left(- 1 + \sqrt{2}\right) x}$

Particular Solution

In order to find a particular solution of the non-homogeneous equation we would look for a solution of the form:

$y = K {e}^{a x}$

Where the constants $K$ is to be determined by direct substitution and comparison:

Differentiating wrt $x$ we get:

$\setminus y ' = a K {e}^{a x}$
$y ' ' = {a}^{2} K {e}^{a x}$

Substituting into the DE [A] we get:

$\left({a}^{2} K {e}^{a x}\right) + 2 \left(a K {e}^{a x}\right) - \left(K {e}^{a x}\right) = {e}^{a x}$

$\therefore K \left({a}^{2} + 2 a - 1\right) = 1$

$\therefore K = \frac{1}{{a}^{2} + 2 a - 1}$

We are also given that the PS is $- \frac{4}{7} {e}^{a x}$

$- \frac{4}{7} {e}^{a x} = K {e}^{a x} \implies K = - \frac{4}{7}$

Allowing us to find $a$ using:

$\frac{1}{{a}^{2} + 2 a - 1} = - \frac{4}{7}$

$\therefore 4 \left({a}^{2} + 2 a - 1\right) = - 7$

$\therefore 4 {a}^{2} + 8 a + 3 = 0$

$\therefore \left(2 a + 1\right) \left(2 a + 3\right) = 0$

$\therefore a = - \frac{3}{2} , - \frac{1}{2}$

We cannot eliminate a solution, thus we are left with two possibilities, Allowing us to write the General Solution $y = {y}_{c} + {y}_{p}$ as:

$y = A {e}^{\left(- 1 - \sqrt{2}\right) x} + B {e}^{\left(- 1 + \sqrt{2}\right) x} - \frac{4}{7} {e}^{- \frac{3}{2} x}$

or

$y = A {e}^{\left(- 1 - \sqrt{2}\right) x} + B {e}^{\left(- 1 + \sqrt{2}\right) x} - \frac{4}{7} {e}^{- \frac{1}{2} x}$