2 years, 6 months ago
Laplace Transformation Problem.
1. Solve for y(s, the Laplace Transform of the solution y(t) to the given initial value problem
y"+y'-y=t3;y(0)=0, y'(0)=0?
y"+y'-y=t3;y(0)=0, y'(0)=0?
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M$1 Answer
Taking the L-transform of each term in equation yields:
L(y'') + L(y') - L(y) = 3*L(t)
I'll put braces around each transform substitution:
{s^2*Y(s) - s*y(0) - y'(0)} + {s*Y(s) - y(0)} - {Y(s)} = 3*{1/s^2}
Substituting in initial value conditions:
{s^2*Y(s) - s*0 - 0} + {s*Y(s) - 0} - {Y(s)} = 3/s^2
Simplifying:
{s^2*Y(s)} + {s*Y(s)} - {Y(s)} = 3/s^2
Getting rid of the braces and factoring Y(s) from each term on the left:
Y(s) * (s^2 + s - 1) = 3/s^2
Solving for Y(s) by dividing out polynomial on left:
Y(s) = 3/{s^2 * (s^2 + s - 1)}
I put the braces around the denominator on the right to make it clear that the whole polynomial lies in the denominator.
This answers your question. If you wanted to compute y(t) from here you would either need partial fraction decomposition or some sort of algebraic gymnastic to form fit the right side into something you would see in the LaPlace transform table.
Good luck!
L(y'') + L(y') - L(y) = 3*L(t)
I'll put braces around each transform substitution:
{s^2*Y(s) - s*y(0) - y'(0)} + {s*Y(s) - y(0)} - {Y(s)} = 3*{1/s^2}
Substituting in initial value conditions:
{s^2*Y(s) - s*0 - 0} + {s*Y(s) - 0} - {Y(s)} = 3/s^2
Simplifying:
{s^2*Y(s)} + {s*Y(s)} - {Y(s)} = 3/s^2
Getting rid of the braces and factoring Y(s) from each term on the left:
Y(s) * (s^2 + s - 1) = 3/s^2
Solving for Y(s) by dividing out polynomial on left:
Y(s) = 3/{s^2 * (s^2 + s - 1)}
I put the braces around the denominator on the right to make it clear that the whole polynomial lies in the denominator.
This answers your question. If you wanted to compute y(t) from here you would either need partial fraction decomposition or some sort of algebraic gymnastic to form fit the right side into something you would see in the LaPlace transform table.
Good luck!
You can leave an optional "tip" with Mahalo's virtual currency, Mahalo Dollars. If you are asking a difficult question that might require some research, or if you'd like a wide variety of feedback, a higher tip often leads to more answers to your question.
M$
Hi Thanks for answering but you misinterpret my question as y'' + y' -y = 3*t instead of y"+y'-y=t3 (this is cube and not multiplication). Please help me to solve by taking it as a cube
Sorry about that. This is an easy fix. Instead of 3*t being L-transformed into 3/s^2 it will be transformed into 3!/s^4 = 6/s^4.
None of the algebra changes, so we can substitute this directly into the final answer:
Y(s) = 6/{s^4*(s^2 + s -1)}
I can see why the question asked you to stop here now. Trying partial fractions on this would be much more tedious and there is no obvious shift to simplify the expression on the right.
Hope this helps!
Hi Thanks for answering but you misinterpret my question as y'' + y' -y = 3*t instead of y"+y'-y=t^3 (this is cube and not multiplication). Please help me to solve by taking it as a cube