First Order Differential Equations

A first order differential equation involves a function y(x) and its first derivative dy/dx. In Further Mathematics, you need to solve several types: separable, linear (using an integrating factor), and use differential equations in modelling real-world problems.

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1 Separable Equations

Definition

A separable equation can be written in the form:

dy/dx = f(x) g(y)

This can be separated as:

(1/g(y)) dy = f(x) dx

Then integrate both sides.

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Worked Example 1

Solve dy/dx = xy, given y(0) = 2.

Solution:

Separate: (1/y) dy = x dx

Integrate: ln|y| = x^2/2 + C

So y = A e^(x^2/2) where A = e^C.

Apply y(0) = 2: 2 = A e^0 = A. So A = 2.

y = 2 e^(x^2/2)

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Worked Example 2

Solve dy/dx = (1 + y^2) / (1 + x^2).

Solution:

Separate: dy/(1 + y^2) = dx/(1 + x^2)

Integrate: arctan y = arctan x + C

So y = tan(arctan x + C).

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Worked Example 3

Solve x dy/dx = y(1 - y), given y(1) = 1/2.

Solution:

Separate: dy / (y(1-y)) = dx / x

Use partial fractions on the left: 1/(y(1-y)) = 1/y + 1/(1-y)

integral of (1/y + 1/(1-y)) dy = integral of 1/x dx

ln|y| - ln|1-y| = ln|x| + C

ln|y/(1-y)| = ln|x| + C

y/(1-y) = Ax where A = plus or minus e^C

Apply y(1) = 1/2: (1/2)/(1/2) = A(1), so A = 1.

y/(1-y) = x, so y = x(1-y) = x - xy, giving y + xy = x, y(1+x) = x.

y = x/(1+x)

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2 Linear First Order Equations and the Integrating Factor

Standard Form

A linear first order ODE has the form:

dy/dx + P(x) y = Q(x)

The Integrating Factor Method

The integrating factor is:

mu(x) = e^(integral of P(x) dx)

Multiply both sides by mu(x):

d/dx [mu(x) y] = mu(x) Q(x)

Then integrate both sides to find y.

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Worked Example 4

Solve dy/dx + y/x = x, for x > 0.

Solution:

Here P(x) = 1/x.

mu(x) = e^(integral of 1/x dx) = e^(ln x) = x

Multiply through by x:

x dy/dx + y = x^2

d/dx [xy] = x^2

Integrate: xy = x^3/3 + C

y = x^2/3 + C/x

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Worked Example 5

Solve dy/dx + 2y = e^(-x), given y(0) = 3.

Solution:

P(x) = 2. mu(x) = e^(2x).

Multiply: e^(2x) dy/dx + 2 e^(2x) y = e^(2x) e^(-x) = e^x

d/dx [e^(2x) y] = e^x

Integrate: e^(2x) y = e^x + C

y = e^(-x) + C e^(-2x)

Apply y(0) = 3: 3 = 1 + C, so C = 2.

y = e^(-x) + 2 e^(-2x)

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Worked Example 6

Solve dy/dx - 3y = sin x.

Solution:

P(x) = -3. mu(x) = e^(-3x).

Multiply: d/dx [e^(-3x) y] = e^(-3x) sin x

Integrate the right side using the formula for integral of e^(ax) sin(bx) dx:

integral of e^(-3x) sin x dx = e^(-3x)(-3 sin x - cos x) / (9 + 1) + C = e^(-3x)(-3 sin x - cos x) / 10 + C

So e^(-3x) y = e^(-3x)(-3 sin x - cos x)/10 + C

y = (-3 sin x - cos x)/10 + C e^(3x)

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3 Exact Equations (Brief Introduction)