What is the general solution of the differential equation? :  (d^2y)/dx^2-dy/dx-2y=4x^2

May 10, 2017

$y = A {e}^{- x} + B {e}^{2 x} - 2 {x}^{2} + 2 x - 3$

Explanation:

There are two major steps to solving Second Order DE's of this form:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 - \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y = 4 {x}^{2}$

1) Find the Complementary Function (CF)
This means find the general solution of the Homogeneous Equation

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 - \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y = 0$

To do this we look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

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

This has two distinct real solutions, $m = - 1 , 2$

And so the solution to the DE is;

$y = A {e}^{- x} + B {e}^{2 x}$ Where $A , B$ are arbitrary constants

-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Verification:

If $y = A {e}^{- x} + B {e}^{2 x}$, then
$y ' = - A {e}^{- x} + 2 B {e}^{2 x}$
$y ' ' = A {e}^{- x} + 4 B {e}^{2 x}$

And so, $y ' ' - y ' - 2 y = A {e}^{- x} + 4 B {e}^{2 x} - \left(- A {e}^{- x} + 2 B {e}^{2 x}\right) - 2 \left(A {e}^{- x} + B {e}^{2 x}\right) = 0$
-+-+-+-+-+-+-+-+-+-+-+-+-+-+

2) Find a Particular Integral* (PI)

This means we need to find a specific solution (that is not already part of the solution to the Homogeneous Equation). As the RHS is a quadratic we try a solution of the quadratic form:

$y = a {x}^{2} + b x + c$

Where $a , b , c$ are constants to be found

If $y = a {x}^{2} + b x + c$ then we have:

$y ' \setminus = 2 a x + b$
$y ' ' = 2 a$

If we substitute into the initial DE we get:

$\setminus \setminus 2 a - \left(2 a x + b\right) - 2 \left(a {x}^{2} + b x + c\right) = 4 {x}^{2}$

$\therefore 2 a - 2 a x - b - 2 a {x}^{2} - 2 b x - 2 c = 4 {x}^{2}$

Equating Coefficients we have

$C o e f \left({x}^{2}\right) : - 2 a = 4 \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies a = - 2$
$C o e f \left({x}^{1}\right) : - 2 a - 2 b = 0 \implies b = 2$
$C o e f \left({x}^{0}\right) : 2 a - b - 2 c = 0 \setminus \setminus \setminus \setminus \implies c = - 3$

So we have found that a Particular Solution is:

$y = - 2 {x}^{2} + 2 x - 3$

3) General Solution (GS)
The General Solution to the DE is then:

GS = CF + PI

Hence The General Solution to the initial DE is

$y = A {e}^{- x} + B {e}^{2 x} - 2 {x}^{2} + 2 x - 3$