Advanced Engineering Mathematics
Lecture 2Separable Differential Equations
LecturerHayder Hassan Abbas
9 Oct.2013Petroleum Engineering Department
School of Chemical and Petroleum EngineeringKoya University
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• A differential equation is an equation involving an unknownfunction and its derivatives.
• The general solution of ODE contains constants of integration,that may be determined by the boundary condition.
• Order of the differential equation is the order of the highestderivatives.
Review of Last Lecture
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Solutions of First-Order DifferentialEquations
• Separable Equations
• Reduction of Separable (Homogeneous Equations)
• Exact Equations
• Linear Equations
• Bernoulli Equations
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Separable Equations
• General SolutionThe solution to the first-order separable differentialequation.
A(x)dx + B(y) dy = 0 is
where c represents an arbitrary constant.
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Integrating the last equation Cdxxfy )(
Example: xy 2' CxCdxxy 22
Where x2+C is general solution of the differential equation
dxxfdy )(
Sometimes an additional condition is given like 3)2( y
that means the function y(x) must pass through a point ]3,2[0 x
123 2 CC
We have obtained a particular solution y(x)=x2 -1
1)( 2 xxy
)(xfdxdy
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Example: Solve the following differential equation:
dydx+ x= 0 .
Rearanging terms, we have
dy= − xdx .
[Here B(y) = 1 and A(x) = -x.]6
The solution is found by integrating (finding the anti-derivative of) both sides:
∫ dy = − ∫ xdx.
y = − x2
2+ c .
Any constant value c can be used.
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Example: Solve the following differential equation:
y dydx+ x= 0 .
Rearanging terms, we have
y dy= − x dx .
[Here B(y) = y and A(x) = -x.]8
Integrating both sides, we have
∫ ydy = −∫ xdx.
y2
2= − x2
2+c .
x2+ y2 = 2 c ≡ c ' .
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Example: Solve the following differential equation:
dydx+ y= 0.
Rearanging terms, we have
dyy =− dx.
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Integrating both sides, we have
∫ dyy = − ∫ dx .
ln y = − x+c .y = e− x+c = c ' e− x .
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1)2
exp(
)2
exp()2
exp(1
2)1ln(
1
)1(
2
22
2
xAy
xACxy
Cxyxdxy
dy
yxxyxdxdy
Separable - Example:SolutionGeneralthefind:1Example
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)sin(tydtdy
Separate:dttdy
y)sin(1
Now integrate:
)cos(
)cos(
)cos()ln(
)sin(1
t
ct
Aeyey
cty
dttdyy
:SolutionGeneralthefind:2Example
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Boundary Conditions
In all of the differential equations that we have solved, thesolution contained a constant c. Generally, any value for cwill work as a solution unless further restrictions areimposed. These restrictions are typically initialconditions or boundary conditions.
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Solutions to the Initial-Value Problem
• can be obtained, as usual, by first usingEquation to solve the differential
A(x)dx B(y)dy 0; y(x0) y0• equation and then applying the initial
condition directly to evaluate c.
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dydx+3y= 1,
Example: Find the solution to
where we impose the boundary condition y(0) = 0.
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y ( x )= e − 3x ∫ e 3x dx+ ce − 3x= 13 + ce− 3x .
We have already found the solution to this equation:
Now if y(0) = 0, then
y ( 0 )= 0= 13+ ce− 0 .
We see that c must be –1/3.
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The solution (which satisfies the initial conditions)becomes
y( x ) = 13− 1
3e− 3x .
This solution will not work for different boundaryconditions.
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dydx+3y= 1,
Example: Find the solution to
where we impose the boundary condition y(0) = 1.
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y ( 0 )= 1= 13+ce− 0 .
Imposing the boundary condition, we have
Thus, for this boundary condition, c=2/3, and the solutionis
y( x )= 13+ 2
3e− 3x .
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Examples:1. Solve : dyxdxydyx 24
Answer:
21
Examples:1. Solve :
yx
dxdy 22
Answer:
22
