Solving Differential Equations by Separating Variables - Mathcentre

community project

mathcentre community project encouraging academics to share maths support resources

All mccp resources are released under an Attribution Non-commerical Share Alike licence

Solving Differential Equations by Separating Variables mccp-dobson-1111

Introduction Suppose we have the first order differential equation dy P (y) = Q(x) dx where Q(x) and P (y) are functions involving x and y only respectively. For example 1 1 dy x−3 dy = 3 or = . y2 dx x y 2 dx x3 We can solve these differential equations using the technique of separating variables.

General Solution By taking the original differential equation dy = Q(x) dx we can solve this by separating the equation into two parts. We move all of the equation involving the y variable to one side and all of the equation involving the x variable to the other side, then we dy can integrate both sides. Although dx is not a fraction, we can intuitively treat it like one to move the ”dx” to the right hand side. So Z Z dy P (y) = Q(x) ⇔ P (y) dy = Q(x) dx. dx P (y)

Example Let us find the general solution of the differential equation dy 1 y2 = 3. dx x 1 = 3 ⇔ y dx x 2 dy

⇔

Z

Z

2

y dy = 2

y dy =

Z

Z

1 dx x3 x−3 dx

y3 −x−2 = + c where c is a constant 3 2 −3 ⇔ y 3 = 2 + 3c 2x r 3 −3 + 3c ⇔y= 2x2

⇔

www.mathcentre.ac.uk

c

Katy Dobson University of Leeds

Alan Slomson University of Leeds

Example To find the general solution of the differential equation

mathcentre community project y 2 (x − 3) dy = dx x3

encouraging academics to share maths support resources

we first need to move the y 2 to are thereleased left hand side of the Non-commerical equation. Then move All mccp resources under an Attribution Sharewe Alike licencethe dx to the right hand side of the equation and integrate both sides. y 2 (x − 3) 1 dy x−3 dy = ⇔ 2 = 3 dx x y dx x3 Z Z 1 x−3 ⇔ dy = dx 2 y x3 Z Z 1 x 3 ⇔ dy = − 3 dx 2 3 y x x Z Z 1 3 ⇔ y −2 dy = − 3 dx 2 x x Z Z ⇔ y −2 dy = x−2 − 3x−3 dx 3x−2 +c ⇔ −y = −x + 2 1 3 −1 =− + 2 +c ⇔ y x 2x 1 1 3 ⇔ = − 2 −c y x 2x 2x 3 2cx2 1 ⇔ = 2− 2− y 2x 2x 2x2 1 2x − 3 − 2cx2 ⇔ = y 2x2 2x2 ⇔y= 2x − 3 − 2cx2 −1

where c is a constant

−1

Exercises Find the general solution of dy = y(1 + ex ) dx

1.

2.

dy x = dx y

3.

dy = 9x2 y dx

4.

4 dy 1 = y 3 dx x

Answers

1.

x+ex +c

y=e

2.

√ y = ± x2 + 2c

3.

3x3 +c

y=e

4.

s

y=±

−2 ln |x| + c

Note that the √± symbol like in Exercise 2 means that the differential equation has two sets of √ solutions, y = x2 + 2c and y = − x2 + 2c.

www.mathcentre.ac.uk

c

Katy Dobson University of Leeds

Alan Slomson University of Leeds