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### Course: Differential equations>Unit 3

Lesson 3: Laplace transform to solve a differential equation

# Using the Laplace transform to solve a nonhomogeneous eq

Solving a non-homogeneous differential equation using the Laplace Transform. Created by Sal Khan.

## Want to join the conversation?

• When Sal was doing the partial fractions he used "As + B" and "Cs + D". I just don't understand why he used 4 constants to solve for instead of just using A and B. A couple of videos ago he used A and B and it worked just fine. Could someone please clarify. thanks
• the terms on top of the fractions depend on whats on the bottom, if the factor on the bottom has a squared, then you need Ax + B on top. If its not squared, its just x, or x and constant, then you only need A on the top. for example if you have:

2/(Xsquared+1)(Xnotsquared+1) then when doing the partial fraction decomposition youll want

= (Ax+B) / (Xsquared+1) and the other fraction is just C/(Xnotsquared+1)
• Will there be videos in the future where you calculate the inverse Laplace transform using residues (complex plane)?
• in the first video here, he said L(y'')=sL(y') -y'(0)
in (Laplace transform to solve an equation part 1)
in this video, he says L(y'')=s^2L(y)-sy(0)-y'(0)
which one is correct or are they both correct?
• They are both correct. Look closely at both equations, and you will see that they are indeed equivalent using the relationship L(y') = sL(y) - y(0). All you have to do is substitute this equation into the first equation you listed and you will get the second.
• I don't understand- did he make a mistake in the first example? He says that the Laplace of y'' is s^2L(y)-s*y(0)-y'(0). But wouldn't it be something more like s^2 * L(y') -s*y'(0)?
(1 vote)
• L(y'') = s*L(y') - y'(0) and L(y') = s*L(y) - y(0) so L(y'') = s*(s*L(y) - y(0)) - y'(0) = s^2L(y) -s*y(0) - y'(0)
• What do you do if the initial conditions are not set equal to zero. Like y(2)=3 and y'(2)=0
• I feel like it should be pointed out that, at least according to some definitions, initial conditions do not have to be given at t = 0. For instance, according to "Differential Equations and Their Applications" by Martin Braun, 4th edition, page 128, initial conditions for a second-order differential equation are of the form y(t_0) = y_0; y'(t_0) = y'_0, where t_0 may be nonzero.
(1 vote)
• Would I be able to solve this DE using the methods shown in the videos about complex and repeated roots and undetermined coefficients?
(1 vote)
• What if the initial conditions are not given at y(0) and y'(0) but at different points instead. Then that means I cannot solve the differential equation through Laplace transforms?
(1 vote)
• No you can, just replace y(0), y'(0), y''(0) and so on with c1, c2, c3, ... respectively. You'll see that you can still solve the DE without knowing the initial conditions, try it out
(1 vote)
• Is it possiple to just solve for the homogenous equation, as in previous videos, and then just add a particular solution in the end? I guess that would simplify the calculations a bit, if possible
(1 vote)
• Hey, this is beautiful! Why did we learn solving homogeneous equations, this is a much nicer way no?
(1 vote)
• does initial condition is necessary to solve the differential equation by laplace transformation?
(1 vote)