Answer to Question #88155 in Differential Equations for Philips

Question

solve the second order differential equation
d
2
y
d
x
2
−
4
y
=
12
x
,
y
(
0
)
=
4
,
y
′
(
0
)
=
1
d^2y/dx^2−4y=12x,y(0)=4,y′(0)=1

Answer 1

The general solution of this nonhomogeneous differential equation can be written as

$$\text{y}\left( x \right)={{y}_{c}}\left( x \right)+{{y}_{p}}\left( x \right)$$

where y_c is the general solution of the complementary equation

$$\frac{{{d}^{2}}y}{d{{x}^{2}}}-4y=0$$

and  y_p is a particular solution of equation

$$\frac{{{d}^{2}}y}{d{{x}^{2}}}-4y=12x$$

First find the general solution of the complementary equation. For this we solve the auxiliary equation

$${{r}^{2}}-4=0\,\,\,=>\,\,{{r}_{1}}=2,\,\,\,{{r}_{2}}=-2$$

So, the solution of the complementary equation is

$${{y}_{c}}={{c}_{1}}{{e}^{2x}}+{{c}_{2}}{{e}^{-2x}}$$

Since the right side of the original equation is a polynomial of degree 1, i.e. 12x, we seek a particular solution of the form

$${{y}_{p}}\left( x \right)=Ax+B$$

Then

$${{{y}'}_{p}}\left( x \right)=A,\,\,\,{{{y}''}_{p}}\left( x \right)=0$$

So, substituting into the given differential equation, we have

$$0-4\left( Ax+B \right)=12x$$

or

$$-4Ax-4B=12x$$

Polynomials are equal when their coefficients are equal. Thus

$$A=-3,\,\,\,B=0$$

Therefore, a particular solution is

$${{y}_{p}}\left( x \right)=-3x$$

and the general solution is

$$y\left( x \right)={{c}_{1}}{{e}^{2x}}+{{c}_{2}}{{e}^{-2x}}-3x$$

Now we find c₁ and c₂. Use the condition y(0)=4

$$y\left( 0 \right)={{c}_{1}}{{e}^{2\cdot 0}}+{{c}_{2}}{{e}^{-2\cdot 0}}-3\cdot 0={{c}_{1}}+{{c}_{2}}=4$$

Find y′(x) and then use the condition y′(0)=1

$${y}'\left( x \right)={{\left( {{c}_{1}}{{e}^{2x}}+{{c}_{2}}{{e}^{-2x}}-3x \right)}^{\prime }}=2{{c}_{1}}{{e}^{2x}}-2{{c}_{2}}{{e}^{-2x}}-3$$

$${y}'\left( 0 \right)=2{{c}_{1}}{{e}^{2\cdot 0}}-2{{c}_{2}}{{e}^{-2\cdot 0}}-3=2{{c}_{1}}-2{{c}_{2}}-3=1$$

From the first equation we get

$${{c}_{2}}=4-{{c}_{1}}$$

and substitute it into the second equation

$$2{{c}_{1}}-2\left( 4-{{c}_{1}} \right)-3=1$$

or

$$4{{c}_{1}}=12$$

Then

$${{c}_{1}}=3,\,\,\,{{c}_{2}}=4-3=1$$

So, the solution of the given equation is

$$y\left( x \right)=3{{e}^{2x}}+{{e}^{-2x}}-3x$$