The general solution of this nonhomogeneous differential equation can be written as
y(x)=yc(x)+yp(x) where yc is the general solution of the complementary equation
dx2d2y−4y=0 and yp is a particular solution of equation
dx2d2y−4y=12x First find the general solution of the complementary equation. For this we solve the auxiliary equation
r2−4=0=>r1=2,r2=−2 So, the solution of the complementary equation is
yc=c1e2x+c2e−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 Then
y′p(x)=A,y′′p(x)=0 So, substituting into the given differential equation, we have
0−4(Ax+B)=12x or
−4Ax−4B=12x Polynomials are equal when their coefficients are equal. Thus
A=−3,B=0 Therefore, a particular solution is
yp(x)=−3x and the general solution is
y(x)=c1e2x+c2e−2x−3x Now we find c1 and c2. Use the condition y(0)=4
y(0)=c1e2⋅0+c2e−2⋅0−3⋅0=c1+c2=4 Find y′(x) and then use the condition y′(0)=1
y′(x)=(c1e2x+c2e−2x−3x)′=2c1e2x−2c2e−2x−3
y′(0)=2c1e2⋅0−2c2e−2⋅0−3=2c1−2c2−3=1 From the first equation we get
c2=4−c1 and substitute it into the second equation
2c1−2(4−c1)−3=1 or
4c1=12 Then
c1=3,c2=4−3=1 So, the solution of the given equation is
y(x)=3e2x+e−2x−3x
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