In Problems 9–20, determine whether the equation is exact.
If it is, then solve it.
The solution is .
The condition for exact is .
This equation is exact.
Therefore, the solution of the differential equation is
Hence the solution is
Variation of Parameters. Here is another procedure for solving linear equations that is particularly useful for higher-order linear equations. This method is called variation of parameters. It is based on the idea that just by knowing the form of the solution, we can substitute into the given equation and solve for any unknowns. Here we illustrate the method for first-order equations (see Sections 4.6 and 6.4 for the generalization to higher-order equations).
(a) Show that the general solution to (20) has the form, where ( is a solution to equation (20) when ,
C is a constant, and for a suitable function v(x). [Hint: Show that we can take and then use equation (8).] We can in fact determine the unknown function by solving a separable equation. Then direct substitution of v in the original equation will give a simple equation that can be solved for v.
Use this procedure to find the general solution to (21) localid="1663920708127" , x > 0 by completing the following steps:
(b) Find a nontrivial solution to the separable equation (22) localid="1663920724944" , localid="1663920736626" .
(c) Assuming (21) has a solution of the formlocalid="1663920777078" , substitute this into equation (21), and simplify to obtain localid="1663920789271" .
d) Now integrate to get localid="1663920800433"
(e) Verify that localid="1663920811828" is a general solution to (21).
94% of StudySmarter users get better grades.Sign up for free