Download 22. Linear Equations

Calculus 2
Lia Vas
Linear Differential Equation
A first order differential equation is linear if it can be written in the form a1 (x)y 0 +a0 (x)y = b(x).
Note that if a1 (x) = 0, the equation is not differential. So, let us assume that the function a1 (x)
is not zero. In this case we can divide the equation with a1 (x) and obtain the form
y 0 + P (x)y = Q(x).
Note that here P = a0 /a1 and Q = b/a1 .
To solve this differential equation you should:
1. Write the equation in the form y 0 + P (x)y = Q(x).
2. Find the integrating factor I(x) = e
R
P (x)dx
and multiply both sides of the equation with it.
3. Note that the left side is the derivative of the product I(x) · y.
4. Integrate both sides. On the left side you will have the product I(x) · y.
5. Solve for y.
Practice Problems. Solve the following equations.
1. y 0 + 2y = 2ex ,
y(0) = 1.
2. y 0 − 2y = x.
3. xy 0 + 2y = x3 .
4. x2 y 0 + xy = 1,
y(1) = 2.
5. xy 0 + 2y = cos x,
y(π) = 0.
Solutions.
0
x
1. For the
R equation y + 2y = 2e , you have that P = 2. Determine the integrating factor as
I = e 2dx = e2x . Multiply the equation by it to get y 0 e2x + 2e2x y = 2ex e2x . Note that the left
side is the derivative of the product ye2x (check: the product rule for ye2x gives you y 0 e2x +2e2x y
which is exactly
the left side). So, the equation becomes (ye2x )0 = 2e3x . Integrate both sides
R
to get ye2x = 2e3x dx ⇒ ye2x = 32 e3x + c. Finally, divide by e2x to get the general solution
y=
2 3x
e +c
3
e2x
= 23 ex + ce−2x .
Using the initial condition y(0) = 1, you have 1 = 23 e0 + ce0 = 23 + c ⇒ c = 31 . Thus the solution
is y = 23 ex + 13 e−2x .
1
2. For the equation y 0 − 2y = x,Ryou have that P = −2. Careful: don’t forget the negative sign.
The integrating factor is I = e −2dx = e−2x . Multiply the equation by it to get y 0 e−2x −2e−2x y =
xe−2x . Note that the left side is the derivative of the product
ye−2x . So, the equation becomes
R
(ye−2x )0 = xe−2x . Integrate both sides to get ye−2x = xe−2x dx. Using the integration by parts
e−2x − 14 e−2x + c. Divide
with u = x and dv = e−2x dx for the right side, obtain that ye−2x = −x
2
by e−2x to get the general solution y =
−x −2x 1 −2x
e
−4e
+c
2
e−2x
=
−x
2
− 41 + ce2x .
3. Careful: before determining P, you have to write the equation in the form y 0 + P y = Q. So,
you need to divide
by x first. Obtain y 0 + x2 y = x2 . This gives you that P = x2 . The integrating
R 2
2
factor is I = e x dx = e2 ln x = eln x = x2 . Careful: don’t cancel e2 ln x as 2x.
Multiply the equation by x2 to get y 0 x2 + 2xy = x4 . Note that the left side is the derivative
of the product yx2 . So, the equation becomes (yx2 )0 = x4 . Integrate both sides to get yx2 =
R
x4 dx =
x5
5
+c⇒y =
x5
+c
5
x2
=
x3
5
+
c
.
x2
4. To write the equation in the form y 0 + P y = Q, you need to divide by x2 first. RObtain
1
y 0 + x1 y = x12 . This gives you that P = x1 . Determine the integrating factor now. I = e x dx =
eln x = x. Multiply the equation by x to get y 0 x + y = x1 . Note that the left side is the
derivative
of the product yx. So, the equation becomes (yx)0 = x1 . Integrate both sides to get
R 1
.
yx = x dx = ln x + c ⇒ y = ln x+c
x
Using the initial condition y(1) = 2, you have 2 =
0+c
1
⇒ c = 2. Thus the solution is y =
R
2
ln x+2
.
x
2
5. Divide by x first to get y 0 + x2 y = cosx x . P = x2 ⇒ I = e xRdx = e2 ln x = eln x = x2 . Multiply by
I to get y 0 x2 + 2xy = x cos x ⇒ (yx2 )0 = x cos x ⇒ yx2 = x cos xdx. Using the integration by
parts with u = x and dv = cos xdx, obtain that yx2 = x sin x + cos x + c. Divide by x2 to get
x+c
= x1 sin x + x12 cos x + xc2 .
the general solution y = x sin x+cos
x2
With y(π) = 0 you have that 0 =
is y = x1 sin x + x12 cos x + x12 .
−1
π2
+ πc2 ⇒ 0 = −1 + c ⇒ c = 1. Thus, the particular solution
2