Ordinary Differential Equations/Nonhomogeneous second order equations:Method of undetermined coefficients

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Consider a differencial equation of the form

Clerarly, this is not homogeneous, as .

So, to solve this, we first proceeed as normal, but assume that the equation is homogeneous; set for now. Then the first part of the solution pans like

.

Now we need to find the particular integral. To do this, make an appropriate substitution that relates to what is. For instance, if , then take substitution . As and are multiples of in this case, you'll simply get a linear equation in . Then just plug the value of in the equation.

Hence the solution is

y = general solution + particular integral.

There is one important caveat which you should be aware though. In the previous example for instance, if the general solution already had , the substitution cannot be , as the particular integral cannot be equal to the general solution. In such cases, you need to take the substitution as .

Example[edit | edit source]

Solve the differential equation


Given that

Solution[edit | edit source]

Take . Then

Hence the general form of the equation becomes

Now, the particular integral has to be found. To do so, we consider RHS: . The substation then becomes . Then and . Then the equation reduces to . Hence . The equation is now

Then
. Then

Hence the final equation is