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Separable Differential Equations Another type separable differential equation can be expressed as the product of a function of x and a function of y. Multiply both sides by dx and divide both sides by y 2 to separate the variables. (Assume y 2 is never zero.) Example 1
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Solving First-Order Differential Equations
• A first-order Diff. Eq. In x and y is separable if it can be written so that all the y-terms are on one side and all the x-terms are on the other
First-Order Differential Equations• A differential equation has variables separable if it
is in one of the following forms:
• Integrating both sides, the general solution will be :
dy f(x)
dx g(y)= OR g(y)dy -
f(x)dx = 0
Cdxxfdyyg )()(
Separable Differential Equations
Another type separable differential equation can be expressed as the product of a function of x and a function of y.
dy g x h ydx
22dy xydx
Multiply both sides by dx and divide both sides by y2 to separate the variables. (Assume y2 is never zero.)
2 2 dy x dxy
2 2 y dy x dx
0h y
Example 1
Separable Differential Equations
Another type of separable differential equation can be expressed as the product of a function of x and a function of y.
dy g x h ydx
Example 1
22dy xydx
2 2 dy x dxy
2 2 y dy x dx
2 2 y dy x dx 1 2y x
21 xy
2
1 yx
2
1y Cx
0h y
Example 2
222 1 xdy x y edx
2
2
1 2 1
xdy x e dxy
Separable differential equation
2
2
1 2 1
xdy x e dxy
2u x2 du x dx
2
11
udy e duy
11 2tan uy C e C
211 2tan xy C e C
21tan xy e C Combined constants of integration
Example 2
222 1 xdy x y edx
21tan xy e C We now have y as an implicit function of x.
We can find y as an explicit function of x by taking the tangent of both sides.
21tan tan tan xy e C
2
tan xy e C
Notice that we can not factor out the constant C, because the distributive property does not work with tangent.
Example 3:Differential equation with initial condition – These are called Initial value problems
Solve the differential equation dy/dx = -x/y given the initial condition y(0) = 2.
• Rewrite the equation as ydy = -xdx• Integrate both sides & solve
• Since y(0) = 2, we get 4 + 0 = C, and therefore x2 + y2 = 4
xdxyyd
y2 + x2 = C where C = 2k
kxy 22
21
21
dxxfyg
dy
dxxfyg
dy
ygxfdxdy
)()(
)()(
)()(
Example 4
Solve: 2,1,12
yxwhenx
ydxdy
Solution to Example 4
cxyy
cxdyyy
xdx
ydy
dxxfyg
dy
ygxfx
ydxdy
ln11ln
21
ln1
11
121
1
)()(
)()(1
2
2
2
2
22
22
2
33
33
33
311
31ln
21
11ln
211ln2,1
xxy
xyxy
xyxy
xyy
c
yycyx