Unformatted text preview: MEC702 Wk 3 Lect 3 Separable ODEs. Modeling. Separable ODEs
Many ODEs can be reduced to the form
g(y)y 0 = f (x) by purely algebraic manipulations.
Then integrate both sides with respect to x to obtain
g(y)dy = g(x)dx + c . This method of solving ODEs is called method of separating variables and the equation above is called
a separable equation, because the variables are now separated.
Example. The ODE y 0 = 1 + y 2 is separable because
= 1 + y 2 =⇒
1 + y2 Hence, ˆ dy
1 + y2
= dx = x+c
= 1 tan(x + c). Example. Mixing Problem (Modeling) Tank contains 1000 gal of water in which initially 100 lb of salt is dissolved. Brine runs in at a rate
of 10 gal/min and each gallon contains 5 lbs of dissolved salt. Mixture in the tank is kept uniform by
stirring. Brine runs out at 10 gal/min. Find the amount of salt in the tank at any time t.
Step 1: Setting up a model. Let y(t) = amount of salt in the tank at time t. Then the time rate of change is
y 0 = Salt inow rate − Salt outow rate from Balance Law.
Our model is the ODE
Step 2: Solution of the model.
ODE is separable, so 5lb × 10gal = 50lb of salt.
= 0.01 or 1% of
10gal brine which is
total brine content in the tank, i.e., 0.01y(t).
y 0 = 50 − 0.0y = −0.01(y − 5000). ˆ ˆ
y − 5000
ln |y − 5000| = −0.01t + c∗
y 5000 + Ae−0.01t . = Initially, the tank contains 100lb of salt so y(0) = 100 is the initial condition. Thus,
100 = 5000 + ce0 =⇒ c = −4900. Hence, the amount of salt in the tank at time t is
y(t) = 5000 − 4900e−0.01t . 2 Reduction to Separable Form
Some ODEs are nonseparable but with careful transformation (i.e., substitution) they become separable. We look at the class of ODEs of the form
y0 = f y
x , where f is any dierential function of xy . Set u = xy . Then
y = ux y 0 = u0 x + u . =⇒ Upon substitution, we have
y0 = u0 x + u =
f (u) − u f y x
f (u) − u
x which is separable!!!
Example. Given 2xyy 0 = y0 = y0 = y 2 − x2
y 2 − x2
1 y 1 x
2 y Now substitute y = ux and y 0 = u0 x + u and then simplify to get
u0 x + u =
u0 x Then = u
−u2 − 1
2u ˆ ˆ
du = −
1 + u2
ln 1 + u2
= − ln |x| + c∗ = ln + c∗
x2 + y 2 = cx
+ y2 =
2 a family of circles passing through the origin centering at the x-axis.
3 (1) Exact ODE's. Integrating Factors. Dierential
Note that for a function y = f (x), its derivative
= f 0 (x)
dx is its total derivative. What about for functions z = f (x, y)? We have partial derivatives
fx (x, y) = ∂f
∂x fy (x, y) = ∂f
∂y and And its total derivative is dened using both of the partial derivatives
df = ∂f
∂y (2) It follows from this that if f (x, y) = c a constant, then du = 0.
Example. If then f = x + x2 y 3 = c,
du = 1 + 2xy 3 dx + 3x2 y 2 dy = 0 or
y0 = dy
1 + 2xy 3
3x2 y 2 an ODE we can solve by going backwards!!! Exactness
A rst-order ODE M (x, y) + N (x, y)y 0 = 0 is also written as
M (x, y)dx + N (x, y)dy = 0 is also called an exact dierential equation if its left side
M (x, y)dx + N (x, y)dy is exact, i.e., it forms exactly the total dierential
df = ∂f
∂y of some function f (x, y). Hence Equation (3) can be rewritten as
df = 0. Then by integration, we obtain a general solution of Equation (3) in the form
f (x, y) = c. 4 (3) Implicit Solution and Contours
f (x, y) = c This is called an implicit solution. (Note that the solution of the form
y = h(x) is an explicit solution.)
The graphs of f (x, y) = c are contour lines. Derivation of the Exactness Condition
Comparing (2) and (3), we see that the ODE is exact when
∂x N= ∂f
∂y but we do not know the function f (x, y) to use these conditions. (Equality of Partial Derivatives) Let f (x, y) be a function with continuous partial derivaand ∂f
∂y . Then Theorem. tives ∂f
∂x Using the theorem from Multivariable Calculus given above, we obtain
∂y which we will use from here onwards as the condition for an exact ODE. Finding f (x, y) for Exact ODE
There are two ways of nding f (x, y) for an Exact ODE:
1. Integrating M = ∂f
∂x with respect to x yields
f (x, y) = M dx + k(y) where k(y) is a hidden function in terms of y which we haven't found yet.
To obtain k(y), we nd ∂f
∂y and equate
∂y N= to nd k0 (y). The integrating k0 (y), we nd the hidden function k(y).
2. Integrating N = ∂f
∂y with respect to y yields
f (x, y) = N dy + h(x) where h(x) is a hidden function in terms of x which we haven't found yet.
To obtain h(x), we nd ∂f
∂x and equate
∂x to nd h0 (x). The integrating h0 (x), we nd the hidden function h(x).
5 (4) Example. Consider 0 = cos(x + y)dx + (3y 2 + 2y + cos(x + y))dy. Step 1: Test for Exactness. We notice from Equation (3) that
M = cos(x + y) N = 3y 2 + 2y + cos(x + y). Then checking for exactness (4):
∂x = − sin(x + y) = − sin(x + y). Hence ODE is exact.
Step 2: Implicit General Solution. By integration, ˆ
f = M dx + k(y)
= cos(x + y)dx + k(y)
sin(x + y) + k(y). By partial dierentiation with respect to y and knowing that it is equal to N .
N = ∂f
= cos(x + y) +
y Therefore, = 3y 2 + 2y = y 3 + y 2 + c∗ . f (x, y) = sin(x + y) + y 3 + y 2 = c. Step 3: Checking an Implicit Solution. By dierentiation, we check f (x, y) = c implicitly to see whether it leads to the given ODE.
df = 0 = ∂f
[cos(x + y)] dx + (3y 2 + 2y + cos(x + y)) dy. 6...
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- Winter '16
- Alveen Chand
- Derivative, dx, dy