Page 45 - Schaum's Outline of Differential Equations

28 SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAR 4

4.15. Solve

This differential equation is homogeneous. Substituting Eqs. (4.6) and (4.7) into it, we obtain

which can be algebraically simplified to

2
The solution to this separable equation is In \x\ - v /2 = c, or equivalently

Substituting v = ylx into (1), we find that the solution to the given differential equation is

4.16. Solve

2
2
2
The solution to the differential equation is given in Problem 3.15 as y = x 2 In x +kx . Applying the initial
2
2
2
2
condition, we obtain (-2) = (I) In (1) + k(l) , or k = 4. (Recall that In 1 = 0.) Thus, the solution to the initial-value
problem is
The negative square root is taken, to be consistent with the initial condition.
4.17. Solve

This differential equation is not separable, but it is homogeneous. Noting the (jtAy)-term in the exponential, we
try the substitution u = xly, which is an equivalent form of (4.9). Rewriting the differential equation as

we have upon using substitutions (4.9) and (4.10) and simplifying

This equation is separable; its solution is

which can be rewritten as

Substituting u = xly into (_/), we obtain the solution of the given differential equation as

4.18. Prove that every solution of Eq. (4.2) satisfies Eq. (4.1).
Rewrite (4.1) as A(x) + B(y)y' = 0. If y(x) is a solution, it must satisfy this equation identically in x; hence,