Unformatted text preview: g variable. This method of doing
separation of variables is similar in spirit to using the differential to do substitutions in integration
problems. The manipulations are easier to carry out, though the mathematical reasoning behind
them is somewhat obscured.
20 2 Differential Equations
Example 2.6: Solve the initial value problem dy
= ( y + 1) x , and y (0) = 4 .
The first step is to produce the general solution of the differential equation. The ODE has
separable variables. We separate the variables by dividing both sides by y + 1 and multiplying by
dx . This leads to
= x dx .
We now integrate both sides, treating each side as if the variable on that side were an independent
The integral dy
= xdx = + C .
y +1 ∫
2 ∫ dy /( y + 1) is evaluated using the substitution u = y + 1 and yields ln y + 1 , ignoring the constant of integration. From this we obtain the implicit form of the solution.
ln y + 1 = x2
2 We solve for y by exponentiation of both sides. This yields, (after writing C for eC )
y + 1 = Ce x 2 /2 . The solution y is unwrapped from the absolute value as we did in Example 2.5, producing the
y = Ce x 2 /2 −1 . Having found the general solution, we can solve the initial value problem. Substituting x = 0 and
y = 4 in the last equation gives 4 = Ce0 − 1 = C − 1 , so C = 5 and the solution of the initial value
problem is y = 5e x 2 /2 − 1 .! Example 2.7: Find the general solution of
y (1) = 4 . dy
21 2 y +1
and determine the solution for which
x 2 Differential Equations
We can separate the variables yielding
2y +1 x
We now integrate both sides. ∫ dy
2 y +1 (2.3) On the left side of (2.3) we have ∫ dy
= ∫ (2 y + 1)−1/ 2 dy .
2 y +1 Making the substitution u = 2 y + 1, du = 2 dy the last integral becomes 1 −1/ 2
∫ u du = u = 2 y + 1
2 (ignoring the constant of integration). Therefore (2.3) evaluates to
2 y + 1 = ln x + C , (2.4) which is the implicit...
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