Differential Equations/First Order Linear 1

From Differential Equations

First Order Differential Equations

What is a linear first order equation?

A linear first order equation is an equation in the form

$\frac{dy}{dx}+P(x)y=Q(x)$.

Linear first order equations are important because they show up frequently in nature and physics, and can be solved by a fairly straight forward method.

Solving First Order Linear Equations

First order linear equation can be solved by making them into exact DEs, as seen here.

Step-by-Step Solution

It is not necessary to derive the method of obtaining the integrating factor every time, as it is always the same. Following the steps below is much easier and less prone to error.

To solve the equation

$\frac{dy}{dx}+P(x)y=Q(x),$

1. Calculate the integrating factor, $e^{\int P(x)dx}$. Drop the constant of integration.
2. Multiply the differential equation by the integrating factor.
3. Due to the manipulation leading to (3) on the page Exact DEs, the LHS multiplied by the integrating factor is equivalent to

   $\frac{d}{dx}{e}^{\int P(x)dx}y$.

4. Integrate both sides with respect to x to get

   $y{e}^{\int P(x)dx}=\int Q{e}^{\int P(x)dx}dx+C$

5. Solve for y by dividing out the x factors.

Example 1: P and Q are Constant

Let's say we have the equation

$\frac{dy}{dx}+my=n$

where n and m are constants. Solve this for y.

Step 1: Find the integrating factor, $e^{\int P(x)dx}$

$\int mdx=mx+C$

$e^{\int P(x)dx}=C{e}^{mx}$

Letting C=1, we get e^mx.

Step 2: Multiply through by integrating factor.

$e^{mx}\frac{dy}{dx}+e^{mx}my=n{e}^{mx}$

Step 3: Recognize that the left hand is $\frac{d}{dx}{e}^{\int P(x)dx}y$, giving

$\frac{d}{dx}{e}^{mx}y=n{e}^{mx}$

Step 4: Integrate both sides w.r.t.x.

$\int (\frac{d}{dx}{e}^{mx}y)dx=\int n{e}^{mx}dx$

$e^{mx}y=\frac{n}{m}{e}^{mx}+C$

Step 5: Solve for y

$y=\frac{n}{m}+\frac{C}{e^{mx}}$

Example 2: P and Q are x

Take the equation

$\frac{dy}{dx}+xy=x$

Solve for y.

Step 1: Find $e^{\int P(x)dx}$

$\int x=\frac{1}{2}{x}^2+C$

$e^{\int P(x)dx}=C{e}^{\frac{1}{2}{x}^2}$

Letting C=1, we get $e^{\frac{1}{2}{x}^2}$

Step 2: Multiply through

$e^{\frac{1}{2}{x}^2}\frac{dy}{dx}+e^{\frac{1}{2}{x}^2}xy=e^{\frac{1}{2}{x}^2}x$

Step 3: Recognize that the left hand is $\frac{d}{dx}{e}^{\int P(x)dx}y$, giving

$\frac{d}{dx}{e}^{\frac{1}{2}{x}^2}y=e^{\frac{1}{2}{x}^2}x$

Step 4: Integrate

$\int (\frac{d}{dx}{e}^{\frac{1}{2}{x}^2}y)dx=\int x{e}^{\frac{1}{2}{x}^2}dx$

$e^{\frac{1}{2}{x}^2}y=e^{\frac{1}{2}{x}^2}+C$

Step 5: Solve for y

$y=1+\frac{C}{e^{\frac{1}{2}{x}^2}}$

Example 3: P and Q are Unrelated

Take the equation

$\frac{dy}{dx}+\frac{1}{x}y=x$

Solve for y.

Step 1: Find $e^{\int P(x)dx}$

$\int \frac{1}{x}=ln(x)+C$

$e^{\int P(x)dx}=C{e}^{ln(x)}=Cx$

Letting C=1, we get x.

Step 2: Multiply through

$x\frac{dy}{dx}+y=x^2$

Step 3: Recognize that the left hand is $\frac{d}{dx}{e}^{\int P(x)dx}y$

$\frac{d}{dx}xy=x^2$

Step 4: Integrate both sides w.r.t.x.

$\int (\frac{d}{dx}xy)dx=\int x^{2}dx$

$xy=\frac{1}{3}{x}^3+C$

Step 5: Solve for y

$y=\frac{1}{3}{x}^2+\frac{C}{x}$

Making Linear Equations from Non-Linear Equations

Sometimes a non-linear equation, which is not solvable like this, can be made linear, and more easily solvable, by applying a substitution.

Example

$y\frac{dy}{dx}+x{y}^2=5x$

Let's make the following substitution:

$v=y^2$

$\frac{dv}{dx}=2y\frac{dy}{dx}$

Plugging in, we get

$\frac{1}{2}\frac{dv}{dx}+xv=5x$

$\frac{dv}{dx}+2xv=10x$

We can then solve as a linear equation in v, using the step-by-step method above:

Step 1: Find the integrating factor:

$e^{\int P(x)dx}$

$e^{\int 2xdx}=e^{x^2+C}$

$e^{\int P(x)dx}=C{e}^{x^2}$

Letting C=1 for convenience, we get $e^{x^2}$ as our integrating factor.

Step 2: Multiply through

$e^{x^2}\frac{dv}{dx}+e^{x^2}xv=10e^{x^2}x$

Step 3: Recognize that the left hand is $\frac{d}{dx}{e}^{\int P(x)dx}v$

$\frac{d}{dx}{e}^{x^2}v=10e^{x^2}x$

Step 4: Integrate both sides w.r.t.x.

$\int (\frac{d}{dx}{e}^{x^2}v)dx=\int 10e^{x^2}xdx$

$e^{x^2}v=5e^{x^2}+C$

Step 5: Solve for v.

$v=5+\frac{C}{e^{x^2}}$

Now that we have v, solve for y.

$v=y^2$

$y^2=5+\frac{C}{e^{x^2}}$

Existence of Solutions

Just like with separable equations, not all initial value problems for linear equations have a solution.

Theorem 1: If P(x) and Q(x) are continuous on an interval I containing the point $x_0$, then the initial value problem has a single unique solution.

This is different from separable equations where the conditions for uniqueness and existance are different - with linear equations, if it exists, it will be unique.