# 1. Introduction

d2ydx2 + ( dydx ) 3 = x7

is an example of an ordinary differential equation (o.d.e.) since it contains only ordinary derivatives such as dydx and not partial derivatives such as yx.

The dependent variable is y while the independent variable is x. An o.d.e. has only one independent variable, whilst a partial differential equation has more than one independent variable.

The above example is a second order equation since the highest order derivative involved is two (the d2ydx2 term).

An o.d.e. is linear when each term has y and its derivatives only appearing to the power one. The appearance of a term involving the product of y and dydx would also make an o.d.e. non-linear.

In the above example, the term (dydx)3 makes the equation non-linear.

The general solution of an nth order o.d.e. has n arbitrary constants that can take any values.

In an initial value problem, one solves an nth order o.d.e. to find the general solution then applies n boundary solutions ("initial values/conditions") to find a particular solution that does not have any arbitrary constants.

# 2. Theory

An ordinary differential equation of the following form:

dy/dx = ƒ(x)

can be solved by integrating both sides with respect to x:

y =    ƒ(x) dx

This technique, called direct integration, can also be applied when the left hand side is a higher order derivative.

In this case, one integrates the equation a sufficient number of times until y is found.

# 3. Exercises

Click on the questions to reveal their solution

Exercise 1:

Show that y = 2 e2x is a particular solution of the ordinary differential equation:
d2y/dx2dy/dx − 2y = 0

Solution:

 We have:   dy/dx = 2⋅2 e2x = 4 e2x and:   d2y/dx2 = 2⋅4 e2x = 8 e2x ∴  d2y/dx2 − dy/dx − 2y = 8 e2x − 4 e2x − 2⋅e2x = (8 − 8) e2x = 0 = RHS

Exercise 2:

Show that y = 7 cos (3x) − 2 sin (2x) is a particular solution of:
d2y/dx2 + 2y = −49 cos (3x) + 4 sin (2x)

Solution:

 We have:   dy/dx = −21 sin (3x) − 4 cos (2x) and:   d2y/dx2 = −63 cos (3x) + 8 sin (2x) ∴ d2y/dx2 + 2y = −63 cos (3x) + 8 sin(2x) + 2 (7 cos (3x) − 2 sin(2x)) = (−63 + 14) cos (3x) + (8 − 4) sin (2x) = −49 cos (3x) + 4 sin(2x)

Note: the equation is second order, so the general solution would have two arbitrary (undetermined) constants.

Notice how similar the particular solution is to the right hand side of the equation. It involves the same functions, but they have different coefficient, i.e.: y is of the form a cos (3x) + b sin (2x),   where a = 7 and b = −2.

Exercise 3:

Show that y = A sin (x) + B cos (x), where A and B are arbitrary constants, is the general solution of:
d2y/dx2 + y = 0

Solution:

 dy/dx = A cos (x) − B sin (x) d2y/dx2 = −A sin (x) − B cos (x) ∴ d2y/dx2 + y = (−A sin (x) − B cos (x)) + (A sin (x) + B cos (x)) = 0

Note: since the differential equation is second order and the solution has two arbitrary constants, this solution is the general solution.

Exercise 4:

Derive the general solution of dy/dx = 2x+3

Solution:

This is an equation of the form dy/dx = ƒ(x), and it can be solved by direct integration.

Integrate both sides with respect to x:

 ∫  dy/dx dx = ∫  (2x + 3)dx i.e.    ∫  dy = ∫  (2x + 3)dx i.e.  y = 2⋅1/2 x2 + 3x + C i.e.  y = x2 + 3x + C

- where C is the (combined) arbitrary constant that results from integrating both sides of the equation. The general solution must have one arbitrary constant since the differential equation is first order.

Exercise 5:

Derive the general solution of d2y/dx2 = − sin (x)

Solution:

This is of the form d2y/dx2 = ƒ(x), so we can solve for y by direct integration.

Integrate both sides with respect to x:

 dy/dx = −   ∫   sin (x)dx = − (− cos x) + A

Integrate again:

 y = sin (x) + Ax + B

- where A,B are the two arbitrary constants of the general solution. Note that the equation is second order.

Exercise 6:

Derive the general solution of d2y/dt2 = a,   where a = constant

Solution:

Integrate both sides with respect to t:

 dy/dt = ∫  a dt i.e.  dy/dt = at + C

Integrate again:

 y = ∫  (at + C)dt y = 1/2 at2 + Ct + D

- where C,D are the two arbitrary constants required for the general solution of the second order differential equation.

Exercise 7:

Derive the general solution of d3y/dt2 = 3x2

Solution:

Integrate both sides with respect to x:

 d2y/dx2 = ∫  3x2dx i.e. d2y/dx2 = 3⋅1/3 x3 + C i.e. d2y/dx2 = x3 + C

Integrate again:

 dy/dx = ∫  (x3 + C) dx i.e. dy/dx = x4/4 + Cx + D

Integrate again:

 y = ∫  (x4/4 + Cx + D) dx i.e.  y = 1/20 x5 + C/2 x2 + Dx + E i.e.  y = 1/20 x5 + C'x2 + Dx + E

where C' (= C2), D and E are the required three arbitrary constants for the general solution of the third order differential equation.

Exercise 8:

Derive the general solution of ex d2y/dx2 = 3

Solution:

Multiplying both sides of the equation by ex gives:

 ex⋅e−x d2y/dx2 = ex⋅3 i.e.  d2y/dx2 = 3 ex

This is now of the form d2y/dx2 = ƒ(x), where ƒ(x) = 3 ex , and the solution y can be found by direct integration.

Integrate both sides with respect to x:

 dy/dx = ∫  (3ex)dx i.e.  dy/dx = 3 ex + C

Integrate again:

 y = ∫  (3 ex + C) dx i.e. y = 3 ex + Cx + D

- where C and D are the two arbitrary constants of the general solution of the second order differential equation.

# 3. Standard Integrals

ƒ(x) ∫ƒ(x) dx ƒ(x) ∫ƒ(x) dx xn xn+1/ n+1 (n ≠ −1) [g(x)]n g'(x) [g(x)]n+1/ n + 1 (n ≠ −1) 1/x ln x g'(x)/g(x) ln g(x) ex ex ax ax/ln a (a > 0) sin x −cos x sinh x cosh x cos x sin x cosh x sinh x tan x − ln cosx tanh x ln cosh x cosec x ln tan  x/2 cosech x ln tanh  x/2 sec x ln sec x + tan x sec x 2 tan−1ex sec2 x tan x sec2 x tanh x cot x ln sin x cot x ln sinh x sin2 x x/2 − sin 2x/4 sinh2 x sinh 2x/4 − x/2 cos2 x x/2 + sin 2x/4 cosh2 x sinh 2x/4 + x/2 1/ a2 + x2 1/ a tan −1  x/ a   (a > 0) √a2 + x2 a2/ 2 [ sinh−1( x/ a ) + x √a2−x2 / a2 ] 1/ a2 − x2 1/ 2a ln  a + x/ a − x    (0 < x < a) √a2−x2 a2/ 2 [ sin−1( x/ a ) + x √a2−x2 / a2 ] 1/ x2 − a2 1/ 2a ln  x − a/ x + a   (x > a > 0) √x2 − a2 a2/ 2 [−cosh−1( x/ a ) + x √x2−a2 / a2 ] 1/ √ a2 + x2 ln  x + √a2 + x2 / a   (a > 0) 1/ √ a2 − x2 sin−1  x/ a     (−a < x < a) 1/ √ x2 − a2 ln  x + √x2 − a2 / a   (x > a > 0)