Solve the Differential equations 2y′′ + 3y′ − 2y = 14x^{2} − 4x − 11, y(0) = 0, y′(0) = 0

Dottie Parra

Dottie Parra

Answered question

2020-11-02

Solve the Differential equations 2y+3y2y=14x24x11,y(0)=0,y(0)=0

Answer & Explanation

Fatema Sutton

Fatema Sutton

Skilled2020-11-03Added 88 answers

We first solve the homogeneous equation
2y"+3y'2y=0
Its characteristic equation is
2r2+3r-2=0r

=3±9+164=-3±54
Thus, r=-2 or r=12 the solution of the homogeneous equation is
yH=C1e-2x+C2ex/2
Now we need to find some particular solution y, a solution such that
2y"p,+3y'p-2yp=14x2-4x-11(1)
We will try with polynomial
yp=Ax2+Bx+C
We have that
y'p=2Ax+B
y"p=2A
So, using (1),
2A+2Ax+B+Ax2+Bx+C=14x2-4x-11
Ax2+(B+2A)x+(2A+B+C)=14x2-4x-11)
So, we have a system of equations
A=14
2A+B=-4
2A+B+C=-11
Thus,
A=14,B=-32,C=-7,
so yp=14x2-32x-7
Finally, the solution of the initial equation is given by
y=yH+yp=C1e-2x+C2ex/2+28x-32
Now use that
0=y(0)=C1+C2-7
and
0=y'(0)=-2C1+12C2-32
So, the system is
C1+C2=7
-2C1+12C2=32
Multiply the first equation by 2 and add it to the second equation:
52C2=46C2=18.4
Plug this into the first equation to get
C1+18.4=7C1=-11.4
Finally, the solution is
y=-11.4e-2x+18.4ex/2+14x2-32x-7

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