# How do you find the general solution of the differential equations y''-4y'=2x^2?

Dec 30, 2016

The General Solution to the DE $y ' ' - 4 y ' = 2 {x}^{2}$ is

$y = A + B {e}^{4 x} - \frac{1}{6} {x}^{3} - \frac{1}{8} {x}^{2} - \frac{1}{16} x$

#### Explanation:

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

$y ' ' - 4 y ' = 2 {x}^{2}$

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

$y ' ' - 4 y ' = 0$

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

As $y ' ' - 4 y ' = 0$, the Axillary Equation is:

${m}^{2} - 4 m + 0 = 0$
$\therefore m \left(m - 4\right) = 0$

This has two distinct real solutions, $m = 0$ and $m = 4$

And so the solution to the DE is;

$\setminus \setminus \setminus \setminus \setminus y = A {e}^{0 x} + B {e}^{4 x}$ Where $A , B$ are arbitrary constants
$\therefore y = A + B {e}^{4 x}$

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Verification:

$y = A + B {e}^{4 x} \implies y ' = B {e}^{4 x}$
$\text{ } \setminus \implies y ' ' = 16 B {e}^{4 x}$
$\therefore y ' ' - 4 y ' = 16 B {e}^{4 x} - 4 \left(4 B {e}^{4 x}\right)$
$\text{ } \setminus = 16 B {e}^{- 4 x} - 16 B {e}^{- 4 x}$
$\text{ } \setminus = 0$

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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).

We look for a particular solution which is a combination of functions on the RHS of the form

$y = a {x}^{3} + b {x}^{2} + c x + d$

In which case, we get:

$y ' \setminus = 3 a {x}^{2} + 2 b x + c$
$y ' ' = 6 a x + 2 b$

If we substitute into the DE $y ' ' - 4 y ' = 2 {x}^{2}$ we get:

$\left(6 a x + 2 b\right) - 4 \left(3 a {x}^{2} + 2 b x + c\right) = 2 {x}^{2}$

Equating Coefficients we get:

${x}^{2} : - 12 a = 2 \implies a = - \frac{1}{6}$
${x}^{1} : 6 a - 8 b = 0 \implies b = - \frac{1}{8}$
${x}^{0} : 2 b - 4 c = 0 \implies c = - \frac{1}{16}$

So we have found that a Particular Solution is:

$y = - \frac{1}{6} {x}^{3} - \frac{1}{8} {x}^{2} - \frac{1}{16} x + d$

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Verification:

$y = - \frac{1}{6} {x}^{3} - \frac{1}{8} {x}^{2} - \frac{1}{16} x + d$
$\text{ } \implies y ' ' = - \frac{1}{2} {x}^{2} - \frac{1}{4} x - \frac{1}{16}$
$\text{ } \implies y ' ' = - x - \frac{1}{4}$
$\therefore y ' ' - 4 y ' = \left(- x - \frac{1}{4}\right) - 4 \left(- \frac{1}{2} {x}^{2} - \frac{1}{4} x - \frac{1}{16}\right)$
$\text{ } \setminus = - x - \frac{1}{4} + 2 {x}^{2} + x + \frac{1}{4}$
$\text{ } \setminus = 2 {x}^{2}$

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General Solution (GS)
The General Solution to the DE is then:
GS = CF + PI

Hence The General Solution to the DE $y ' ' - 4 y ' = 2 {x}^{2}$ is

$y = A + B {e}^{4 x} - \frac{1}{6} {x}^{3} - \frac{1}{8} {x}^{2} - \frac{1}{16} x + d$
$y = A + B {e}^{4 x} - \frac{1}{6} {x}^{3} - \frac{1}{8} {x}^{2} - \frac{1}{16} x$ where A is arbitary