Professor Tambuwal, thank you for an excellent video/lecture on Using Laplace Transform to solve Second Order Homogeneous Differential Equations. There are multiple ways to solve Second Order Homogeneous Differential Equations.
What could be the cause if at the level of deriving the partial fraction, where the constants are A,B and C, B and C keep turning out to be zero, does it mean the equation has a problem or my solving is wrong (it was from a different equation)!?
If you know the Laplace transform for y, let's call it capital Y, then to differentiate, you multiply by s and subtract y_0. Multiply by s^2 and subtract (y0' + s*y0) to double differentiate. For this example: y" + 5*y' + 6*y = 0 Consider general initial conditions, y(0) = u, and y'(0) = v. The general solution for Y is: Y = (v + 3*u)/(s + 2) - (v + 2*u)/(s + 3) And its inverse Laplace time domain solution is: y = (3*u + v)*e^(-2*t) - (2*u + v)*e^(-3*t) Multiply by s^2 to differentiate, and then subtract applicable terms for the initial conditions: L{y"} = s^2*Y - u*s - v L{y"} = (v + 3*u)*s^2/(s + 2) - (v + 2*u)*s^2/(s + 3) - u*s - v Change to common denominators: L{y"} = -(6*s*u + 5*s*v + 6*v)/((s + 2)*(s + 3)) Partial fractions: -(6*s*u + 5*s*v + 6*v)/((s + 2)*(s + 3)) = A/(s + 2) + B/(s + 3) Result after using Heaviside coverup to find A & B: L{y"} = (12*u + 4*v)/(s + 2) + (18*u + 9*v)/(s + 3) Which means: y" = (12*u + 4*v)*e^(-2*t) - (18*u + 9*v)*e^(-3*t) Compare with solution for y: y = (3*u + v)*e^(-2*t) - (2*u + v)*e^(-3*t) Observe that y" compares to y, by multiplying each term by the square of the coefficient on t in the exponent. It quadruples (i.e. (-2)^2) the first coefficient, and multiplies the second coefficient by 9, i.e. (-3)^2. Just as we'd expect would happen in a second derivative of exponential functions.
someone said your answer is wrong because at 2mins 35sec you jumped a step in opening the bracket, but you are 100% right overall. Thank you for the video. I got similar question to solve tomorrow
Professor Tambuwal, thank you for an excellent video/lecture on Using Laplace Transform to solve Second Order Homogeneous Differential Equations. There are multiple ways to solve Second Order Homogeneous Differential Equations.
I now understand this method, thanks so much sir
Thank you, very helpful
What could be the cause if at the level of deriving the partial fraction, where the constants are A,B and C, B and C keep turning out to be zero, does it mean the equation has a problem or my solving is wrong (it was from a different equation)!?
Thank you sir
Keep up bro
Thank you
Please help me explain how to solve laplace transform of y’’ on its own. Thanks.
If you know the Laplace transform for y, let's call it capital Y, then to differentiate, you multiply by s and subtract y_0. Multiply by s^2 and subtract (y0' + s*y0) to double differentiate.
For this example:
y" + 5*y' + 6*y = 0
Consider general initial conditions, y(0) = u, and y'(0) = v.
The general solution for Y is:
Y = (v + 3*u)/(s + 2) - (v + 2*u)/(s + 3)
And its inverse Laplace time domain solution is:
y = (3*u + v)*e^(-2*t) - (2*u + v)*e^(-3*t)
Multiply by s^2 to differentiate, and then subtract applicable terms for the initial conditions:
L{y"} = s^2*Y - u*s - v
L{y"} = (v + 3*u)*s^2/(s + 2) - (v + 2*u)*s^2/(s + 3) - u*s - v
Change to common denominators:
L{y"} = -(6*s*u + 5*s*v + 6*v)/((s + 2)*(s + 3))
Partial fractions:
-(6*s*u + 5*s*v + 6*v)/((s + 2)*(s + 3)) = A/(s + 2) + B/(s + 3)
Result after using Heaviside coverup to find A & B:
L{y"} = (12*u + 4*v)/(s + 2) + (18*u + 9*v)/(s + 3)
Which means:
y" = (12*u + 4*v)*e^(-2*t) - (18*u + 9*v)*e^(-3*t)
Compare with solution for y:
y = (3*u + v)*e^(-2*t) - (2*u + v)*e^(-3*t)
Observe that y" compares to y, by multiplying each term by the square of the coefficient on t in the exponent. It quadruples (i.e. (-2)^2) the first coefficient, and multiplies the second coefficient by 9, i.e. (-3)^2. Just as we'd expect would happen in a second derivative of exponential functions.
It is wrong, The answer is not correct.
produce a Video for the correct answer
U don't say it's wrong like that as a mathematician. U show it's wrong by producing correct answer
@@abdulkahin8270 I don't care about his rude comment, let him prove me wrong. I'm ready for his challenge.
It's bcz of u that I understood o.d.e and matrices likes of cofactor of a matrix. I say thank u bro
someone said your answer is wrong because at 2mins 35sec you jumped a step in opening the bracket, but you are 100% right overall. Thank you for the video. I got similar question to solve tomorrow