Page 54 - Schaum's Outline of Differential Equations

CHAP. 5] EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 37

2
The solution to the initial-value problem is y t + y = 2, in implicit form. Solving for y directly, using the quadratic
formula, we have

where the negative sign in front of the radical was chosen to be consistent with the given initial condition.

5.13.

This differential equation in standard form has the differential form of Problem 5.9. Its solution is given in (2)
2
2
of Problem 5.10 as x - xt = c 3. Using the initial condition x = 3 when t = 2, we obtain (3) - 3(2) = c 3, or c 3 = 3. The
2
solution to the initial-value problem is x + xt = 3, in implicit form. Solving for x directly, using the quadratic formula,
we have

where the positive sign in front of the radical was chosen to be consistent with the given initial condition.

5.14. Determine whether -IIx 2 is an integrating factor for the differential equation y dx - x dy = 0.
2
It was shown in Problem 5.3 that the differential equation is not exact. Multiplying it by —llx , we obtain

Equation (1) has the form of Eq. (5.1) with M(x, y) = -ylx 2 and N(x, y) = llx. Now

so (1) is exact, which implies that -llx 2 is an integrating factor for the original differential equation.

5.15. Solve y dx - x dy =0
Using the results of Problem 5.14, we can rewrite the given differential equation as

which is exact. Equation (1) can be solved using the steps described in Eqs. (5.4) through (5.6).
Alternatively, we note from Table 5-1 that (1) can be rewritten as d(ylx) = 0. Hence, by direct integration, we
have ylx = c, or y = ex, as the solution.

5.16. Determine whether —ll(xy) is also an integrating factor for the differential equation defined in Problem 5.14.
Multiplying the differential equation y dx — x dy = 0 by —ll(xy), we obtain

Equation (1) has the form of Eq. (5.1) with M(x, y) = —llx and N(x, y) = lly. Now

so (1) is exact, which implies that —llxy is also an integrating factor for the original differential equation.