1. Lesson 10: The non-homogeneous case
Find the general solutions of the following differential equations
Exercise 10.1.
y 00 − 3y 0 + 2y = sin x
Solution. Let us first find the solution of the homogeneous equation
y 00 − 3y 0 + 2y = 0.
Considering the algebraic equation
m2 − 3m + 2 = 0,
we have the roots m = 1 and m = 2. Therefore, there is the root
c1 ex + c2 e2x
of this equation. Let us now find the partial solution of the non-homogeneous
differential equation, assuming that it is
f (x) = a1 sin x + a2 cos x.
Our task is to find constants a1 and a2 . By differentiating we obtain
f 0 (x) = a1 cos x − a2 sin x,
f 00 (x) = −a1 sin x − a2 cos x.
Thus, we have the equations
−a1 − 3a2 + 2a1 = 1,
−a2 + 3a1 + 2a2 = 0,
or
a1 + 3a2 = 1,
a2 − 3a1 = 0
We obtain a1 = 0.1, a2 = 0.3. Therefore, the final solution of the equation is
y = 0.1 sin x + 0.3 cos x + c1 ex + c2 e2x .
Exercise 10.2.
y 00 + 2y 0 + 2y = 1 + x2 .
Solution. Let us first find the solution of the homogeneous equation
y 00 + 2y 0 + 2y = 0
Consider the associated algebraic equation
m2 + 2m + 2 = 0.
The roots are m1 = −1+i, m2 = −1−i. Therefore the solution of this homogeneous
equation is
e−x (c1 cos x + c2 sin x).
Let us now find a partial solution of non-homogeneous equation in the form
f (x) = a0 x2 + a1 x + a2 .
We have
f 0 (x) = 2a0 x + a1 ,
f 00 (x) = 2a0 .
1
2
Therefore,
2a0 + 2a1 + 2a2 = 1,
4a0 + 2a1 = 0,
2a0 = 1.
We obtain a0 = 0.5, a1 = −1, a2 = 1. The final solution is
y = 1 − x + 0.5x2 + e−x (c1 cos x + c2 sin x).
Exercise 10.3.
y 00 + 4y 0 + 4y = x − 2e2x .
Solution. Let us first find the solution of the homogeneous equation
y 00 + 4y 0 + 4y = 0.
The associated algebraic equation is
m2 + 4m + 4 = 0.
The roots of this equation are m1,2 = −2. Therefore the solution of homogeneous
equation is
c1 e−2x + c2 xe−2x .
Then, the partial solution of the non-homogeneous equation has the following representation
f (x) = a0 x + a1 + a2 e2x .
Then,
f 0 (x) = a0 + 2a2 e2x ,
f 00 (x) = 4a2 e2x .
We have
4a2 + 8a2 + 4a2 = −2
That is
1
a2 = − .
8
Next,
4a0 = 1,
1
a0 = ,
4
and
4a0 + 4a1 = 0,
1
a1 = − .
4
The final solution is
1 x 1
y = − + − e2x + c1 e−2x + c2 xe−2x .
4 4 8
Exercise 10.4.
y 00 + 4y 0 + 8y = x − ex .
Exercise 10.5.
y 00 − 6y 0 + 9y = ex sin x.
Exercise 10.6.
y 00 − 4y 0 + 3y = x3 .
Exercise 10.7.
8y 00 + 4y 0 + y = sin x − 2 cos x.
3
Exercise 10.8.
4y 00 + 4y 0 + y = ex − 2 cos 2x.
Exercise 10.9.
y 00 + y = 3 cos x.
Solution. Let us first find the solution of the homogeneous equation
y 00 + y = 0.
The roots of associate algebraic equation
m2 + 1 = 0
are m1,2 = ±i, and the solution of homogeneous equation is
c1 cos x + c2 sin x.
Let us now find the partial solution of the non-homogeneous differential equation
as
f (x) = a1 x cos x + a2 x sin x.
We have
f 0 (x) = −a1 x sin x + a2 x cos x + a1 cos x + a2 sin x,
f 00 (x) = −a1 x cos x − a2 x sin x − a1 cos x − a2 sin x − a1 sin x + a2 cos x.
Therefore,
2a2 = 3,
−2a1 = 0,
and
3
a2 = ,
2
a1 = 0.
The final solution of the non-homogeneous differential equation is
3
y = x sin x + c1 cos x + c2 sin x.
2
Exercise 10.10.
y 00 + y = ex + 3 cos x.
Exercise 10.11.
y 00 − 3y 0 + 2y = 2 + ex .
Solution. Let us first find the solution of the homogeneous equation
y 00 − 3y 0 + 2y = 0.
The roots of associate algebraic equation
m2 − 3m + 2 = 0
are m1 = 1, m2 = 2, and the solution of homogeneous equation is
c1 ex + c2 e2x .
Let us now find the partial solution of the non-homogeneous differential equation
as
f (x) = a0 x2 + a1 x + a2 + a3 xex .
We have
f 0 (x) = 2a0 x + a1 + a3 xex + a3 ex ,
f 00 (x) = 2a0 + a3 xex + 2a3 ex .
4
Therefore,
a3 = −1.
Next,
2a0 − 3a1 + 2a2 = 2,
a0 = 0,
2a0 + a1 = 0
That is
a1 = 0,
a2 = 1.
The solution of the nonhomogeneous differential equation is
y = 1 − xex + c1 ex + c2 e2x .
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