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Direct Integration |
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PPLATO / Tutorials / Differential Equations |
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1. Introduction
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 ∂y∂x.
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:
can be solved by integrating both sides with respect to x:
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:
Solution:
| We have: dydx | = | 2⋅2 e2x = 4 e2x |
| and: d2ydx2 | = | 2⋅4 e2x = 8 e2x |
| ∴ d2ydx2 − dydx − 2y | = | 8 e2x − 4 e2x − 2⋅e2x |
| = | (8 − 8) e2x | |
| = | 0 | |
| = | RHS |
Exercise 2:
Solution:
| We have: dydx | = | −21 sin (3x) − 4 cos (2x) |
| and: d2ydx2 | = | −63 cos (3x) + 8 sin (2x) |
| ∴ d2ydx2 + 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:
Solution:
| dydx | = | A cos (x) − B sin (x) |
| d2ydx2 | = | −A sin (x) − B cos (x) |
| ∴ d2ydx2 + 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:
Solution:
This is an equation of the form dydx = ƒ(x), and it can be solved by direct integration.
Integrate both sides with respect to x:
| ∫ dydx dx | = | ∫ (2x + 3)dx |
| i.e. ∫ dy | = | ∫ (2x + 3)dx |
| i.e. y | = | 2⋅12 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:
Solution:
This is of the form d2ydx2 = ƒ(x), so we can solve for y by direct integration.
Integrate both sides with respect to x:
| dydx | = | − ∫ 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:
Solution:
Integrate both sides with respect to t:
| dydt | = | ∫ a dt |
| i.e. dydt | = | at + C |
Integrate again:
| y | = | ∫ (at + C)dt |
| y | = | 12 at2 + Ct + D |
- where C,D are the two arbitrary constants required for the general solution of the second order differential equation.
Exercise 7:
Solution:
Integrate both sides with respect to x:
| d2ydx2 | = | ∫ 3x2dx |
| i.e. d2ydx2 | = | 3⋅13 x3 + C |
| i.e. d2ydx2 | = | x3 + C |
Integrate again:
| dydx | = | ∫ (x3 + C) dx |
| i.e. dydx | = | x44 + Cx + D |
Integrate again:
| y | = | ∫ (x44 + Cx + D) dx |
| i.e. y | = | 120 x5 + C2 x2 + Dx + E |
| i.e. y | = | 120 x5 + C'x2 + Dx + E |
where C' (= C⁄2), D and E are the required three arbitrary constants for the general solution of the third order differential equation.
Exercise 8:
Solution:
Multiplying both sides of the equation by ex gives:
| ex⋅e−x d2ydx2 | = | ex⋅3 |
| i.e. d2ydx2 | = | 3 ex |
This is now of the form d2ydx2 = ƒ(x), where ƒ(x) = 3 ex , and the solution y can be found by direct integration.
Integrate both sides with respect to x:
| dydx | = | ∫ (3ex)dx |
| i.e. dydx | = | 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) | |
| 1x | ln x | g'(x)g(x) | ln g(x) | |
| ex | ex | ax | axln 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 x2 | cosech x | ln tanh x2 | |
| 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 | x2 − sin 2x4 | sinh2 x | sinh 2x4 − x2 | |
| cos2 x | x2 + sin 2x4 | cosh2 x | sinh 2x4 + x2 | |
| 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) |