Solution of First Order Linear Differential Equations
You might like to read about Differential Equations and Separation of Variables first!
A Differential Equation is an equation with a function and one or more of its derivatives:
Example: an equation with the function y and its
derivative
\frac{dy}{dx}
Here we will look at solving a special class of Differential Equations called First Order Linear Differential Equations
First Order
They are "First Order" when there is only \frac{dy}{dx} , not \frac{d^{2}y}{dx^{2}} or \frac{d^{3}y}{dx^{3}} etc
Linear
A first order differential equation is linear when it can be made to look like this:
dy | + P(x)y = Q(x) |
dx |
Where P(x) and Q(x) are functions of x.
To solve it there is a special method:
- We invent two new functions of x, call them u and v, and say that y=uv.
- We then solve to find u, and then find v, and tidy up and we are done!
And we also use the derivative of y=uv (see Derivative Rules (Product Rule) ):
dy | = u | dv | + v | du |
dx | dx | dx |
Steps
Here is a step-by-step method for solving them:
- 1. Substitute y = uv, and
dy = u dv + v du dx dx dx dy + P(x)y = Q(x) dx - 2. Factor the parts involving v
- 3. Put the v term equal to zero (this gives a differential equation in u and x which can be solved in the next step)
- 4. Solve using separation of variables to find u
- 5. Substitute u back into the equation we got at step 2
- 6. Solve that to find v
- 7. Finally, substitute u and v into y = uv to get our solution!
Let's try an example to see:
Example: Solve this:
\frac{dy}{dx} − \frac{y}{x} = 1
First, is this linear? Yes, as it is in the form
\frac{dy}{dx}
+ P(x)y = Q(x)
where P(x) = −
\frac{1}{x}
and Q(x) = 1
So let's follow the steps:
Step 1: Substitute y = uv, and \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}
So this: | \frac{dy}{dx} − \frac{y}{x} = 1 | |
Becomes this: | u \frac{dv}{dx} + v \frac{du}{dx} − \frac{uv}{x} = 1 |
Step 2: Factor the parts involving v
Factor v: | u \frac{dv}{dx} + v( \frac{du}{dx} − \frac{u}{x} ) = 1 |
Step 3: Put the v term equal to zero
v term = zero: | \frac{du}{dx} − \frac{u}{x} = 0 | |
So: | \frac{du}{dx} = \frac{u}{x} |
Step 4: Solve using separation of variables to find u
Separate variables: | \frac{du}{u} = \frac{dx}{x} | |
Put integral sign: | ∫ \frac{du}{u} = ∫ \frac{dx}{x} | |
Integrate: | ln(u) = ln(x) + C | |
Make C = ln(k): | ln(u) = ln(x) + ln(k) | |
And so: | u = kx |
Step 5: Substitute u back into the equation at Step 2
(Remember v term equals 0 so can be ignored): | kx \frac{dv}{dx} = 1 |
Step 6: Solve this to find v
Separate variables: | k dv = \frac{dx}{x} | |
Put integral sign: | ∫ k dv = ∫ \frac{dx}{x} | |
Integrate: | kv = ln(x) + C | |
Make C = ln(c): | kv = ln(x) + ln(c) | |
And so: | kv = ln(cx) | |
And so: | v = \frac{1}{k} ln(cx) |
Step 7: Substitute into y = uv to find the solution to the original equation.
y = uv: | y = kx \frac{1}{k} ln(cx) | |
Simplify: | y = x ln(cx) |
And it produces this nice family of curves:
y = x ln(cx)
for various values of c
What is the meaning of those curves? They are the solution to the equation \frac{dy}{dx} − \frac{y}{x} = 1
In other words:
Anywhere on any of those curves
the slope minus
\frac{y}{x}
equals 1
Let's check a few points on the c=0.6 curve:
Estmating off the graph (to 1 decimal place):
Point | x | y | Slope (\frac{dy}{dx}) | \frac{dy}{dx} − \frac{y}{x} |
---|---|---|---|---|
A | 0.6 | −0.6 | 0 | 0 − \frac{−0.6}{0.6} = 0 + 1 = 1 |
B | 1.6 | 0 | 1 | 1 − \frac{0}{1.6} = 1 − 0 = 1 |
C | 2.5 | 1 | 1.4 | 1.4 − \frac{1}{2.5} = 1.4 − 0.4 = 1 |
Why not test a few points yourself? You can plot the curve here.
Perhaps another example to help you? Maybe a little harder?
Example: Solve this:
\frac{dy}{dx} − \frac{3y}{x} = x
First, is this linear? Yes, as it is in the form
\frac{dy}{dx}
+ P(x)y = Q(x)
where P(x) = −
\frac{3}{x}
and Q(x) = x
So let's follow the steps:
Step 1: Substitute y = uv, and \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}
So this: | \frac{dy}{dx} − \frac{3y}{x} = x | |
Becomes this: | u \frac{dv}{dx} + v \frac{du}{dx} − \frac{3uv}{x} = x |
Step 2: Factor the parts involving v
Factor v: | u \frac{dv}{dx} + v( \frac{du}{dx} − \frac{3u}{x} ) = x |
Step 3: Put the v term equal to zero
v term = zero: | \frac{du}{dx} − \frac{3u}{x} = 0 | |
So: | \frac{du}{dx} = \frac{3u}{x} |
Step 4: Solve using separation of variables to find u
Separate variables: | \frac{du}{u} = 3 \frac{dx}{x} | |
Put integral sign: | ∫ \frac{du}{u} = 3 ∫ \frac{dx}{x} | |
Integrate: | ln(u) = 3 ln(x) + C | |
Make C = −ln(k): | ln(u) + ln(k) = 3ln(x) | |
Then: | uk = x^{3} | |
And so: | u = \frac{x^{3}}{k} |
Step 5: Substitute u back into the equation at Step 2
(Remember v term equals 0 so can be ignored): | ( \frac{x^{3}}{k} ) \frac{dv}{dx} = x |
Step 6: Solve this to find v
Separate variables: | dv = k x^{−2} dx | |
Put integral sign: | ∫ dv = ∫ k x^{−2} dx | |
Integrate: | v = −k x^{−1} + D |
Step 7: Substitute into y = uv to find the solution to the original equation.
y = uv: | y = \frac{x^{3}}{k} ( −k x^{−1} + D ) | |
Simplify: | y = −x^{2} + \frac{D}{k} x^{3} | |
Replace D/k with a single constant c: | y = c x^{3 }− x^{2} |
And it produces this nice family of curves:
y = c
x^{3 }− x^{2}
for various values of c
And one more example, this time even harder:
Example: Solve this:
\frac{dy}{dx} + 2xy= −2x^{3}
First, is this linear? Yes, as it is in the form
\frac{dy}{dx}
+ P(x)y = Q(x)
where P(x) = 2x and Q(x) = −2x^{3}
So let's follow the steps:
Step 1: Substitute y = uv, and \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}
So this: | \frac{dy}{dx} + 2xy= −2x^{3} | |
Becomes this: | u \frac{dv}{dx} + v \frac{du}{dx} + 2xuv = −2x^{3} |
Step 2: Factor the parts involving v
Factor v: | u \frac{dv}{dx} + v( \frac{du}{dx} + 2xu ) = −2x^{3} |
Step 3: Put the v term equal to zero
v term = zero: | \frac{du}{dx} + 2xu = 0 |
Step 4: Solve using separation of variables to find u
Separate variables: | \frac{du}{u} = −2x dx | |
Put integral sign: | ∫ \frac{du}{u} = −2 ∫ x dx | |
Integrate: | ln(u) = −x^{2} + C | |
Make C = −ln(k): | ln(u) + ln(k) = −x^{2} | |
Then: | uk = e^{−x2} | |
And so: | u = \frac{e^{−x2}}{k} |
Step 5: Substitute u back into the equation at Step 2
(Remember v term equals 0 so can be ignored): | ( \frac{e^{−x2}}{k} ) \frac{dv}{dx} = −2x^{3} |
Step 6: Solve this to find v
Separate variables: | dv = −2k x^{3} e^{x2} dx | |
Put integral sign: | ∫ dv = ∫ −2k x^{3} e^{x2} dx | |
Integrate: | v = oh no! this is hard! |
Let's see ... we can integrate by parts... which says:
∫ RS dx = R ∫ S dx − ∫ R' ( ∫ S dx) dx
(Side Note: we use R and S here, using u and v could be confusing as they already mean something else here.)
Choosing R and S is very important, this is the best choice we found:
- R = −x^{2} and
- S = 2x e^{x2}
So let's go:
First pull out k: | v = k ∫ −2x^{3} e^{x2} dx | |
R = −x^{2} and S = 2x e^{x2}: | v = k ∫ (−x^{2})(2xe^{x2}) dx | |
Now integrate by parts: | v = kR ∫ S dx − k ∫ R' ( ∫ S dx) dx | |
Put in R = −x^{2} and S = 2x e^{x2} And also R' = −2x and ∫ S dx = e^{x2} |
||
So it becomes: | v = −kx^{2} ∫ 2x e^{x2} dx − k ∫ −2x (e^{x2}) dx | |
Now Integrate: | v = −kx^{2} e^{x2} + k e^{x2} + D | |
Simplify: | v = ke^{x2} (1−x^{2}) + D |
Step 7: Substitute into y = uv to find the solution to the original equation.
y = uv: | y = \frac{e^{−x2}}{k} ( ke^{x2} (1−x^{2}) + D ) | |
Simplify: | y =1 − x^{2} + ( \frac{D}{k})e^{−}^{x2} | |
Replace D/k with a single constant c: | y = 1 − x^{2} + c e^{−}^{x2} |
Done!