I solved the problem using superposition and noticed that it gives you the value of c1=-6. Is it still considered a general solution both ways? Is one more correct? I am assuming it depends how the professor wants it done and what he is looking for
By "general solution", what they mean is the solution that applies regardless of the initial conditions. For this problem, the general solution is: y(t) = C1*e^(3 t) + C2*e^(-3 t) - 6 Or any equivalent expression that gives the same information. You are solving the original differential equation, satisfying both the LHS equal to zero, and the added term to make the LHS equal to the RHS. So this means two out of three of your undetermined coefficients would remain undetermined, and it is the two for the complimentary (or homogeneous) portion of the DiffEQ. In general, there will be n qty undetermined coefficients in the general solution for an nth order DiffEQ.
Amazing explanation. Thanks a lot.
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Do you do linear algebra
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Great Video as always! I love how many applications differential equations have, great fun.
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I solved the problem using superposition and noticed that it gives you the value of c1=-6. Is it still considered a general solution both ways? Is one more correct? I am assuming it depends how the professor wants it done and what he is looking for
By "general solution", what they mean is the solution that applies regardless of the initial conditions. For this problem, the general solution is:
y(t) = C1*e^(3 t) + C2*e^(-3 t) - 6
Or any equivalent expression that gives the same information.
You are solving the original differential equation, satisfying both the LHS equal to zero, and the added term to make the LHS equal to the RHS. So this means two out of three of your undetermined coefficients would remain undetermined, and it is the two for the complimentary (or homogeneous) portion of the DiffEQ. In general, there will be n qty undetermined coefficients in the general solution for an nth order DiffEQ.
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