Question #88155
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
1
Expert's answer
2019-04-17T03:50:30-0400

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

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

where yc is the general solution of the complementary equation

d2ydx24y=0\frac{{{d}^{2}}y}{d{{x}^{2}}}-4y=0

and  yp is a particular solution of equation

d2ydx24y=12x\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

r24=0=>r1=2,r2=2{{r}^{2}}-4=0\,\,\,=>\,\,{{r}_{1}}=2,\,\,\,{{r}_{2}}=-2

So, the solution of the complementary equation is

yc=c1e2x+c2e2x{{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

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

Then

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

So, substituting into the given differential equation, we have

04(Ax+B)=12x0-4\left( Ax+B \right)=12x

or

4Ax4B=12x-4Ax-4B=12x

Polynomials are equal when their coefficients are equal. Thus

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

Therefore, a particular solution is

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

and the general solution is

y(x)=c1e2x+c2e2x3xy\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(0)=c1e20+c2e2030=c1+c2=4y\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(x)=(c1e2x+c2e2x3x)=2c1e2x2c2e2x3{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(0)=2c1e202c2e203=2c12c23=1{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

c2=4c1{{c}_{2}}=4-{{c}_{1}}

and substitute it into the second equation

2c12(4c1)3=12{{c}_{1}}-2\left( 4-{{c}_{1}} \right)-3=1

or

4c1=124{{c}_{1}}=12

Then

c1=3,c2=43=1{{c}_{1}}=3,\,\,\,{{c}_{2}}=4-3=1

So, the solution of the given equation is

y(x)=3e2x+e2x3xy\left( x \right)=3{{e}^{2x}}+{{e}^{-2x}}-3x


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