Use the Laplace transform to solve y^(4)-4y'''+6y"-4y'+y=0; y(0)=0, y'(0)=1, y"(0)=0, y'''(0)=1.
Answer: y=te^t-t^2*e^t+(2/3)t^3*e^t
L(y^(4))-4L(y''')+6L(y")-4L(y')+L(y)=L(0)
But I don't know what's L(y^(4)) and L(y''').
Use the Laplace transform to solve y^(4)-4y'''+6y"-4y'+y=0; y(0)=0, y'(0)=1, y"(0)=0, y'''(0)=1.
Answer: y=te^t-t^2*e^t+(2/3)t^3*e^t
L(y^(4))-4L(y''')+6L(y")-4L(y')+L(y)=L(0)
But I don't know what's L(y^(4)) and L(y''').
This si the amswer to your last question about (s^{2} -4s +7)/(s-1)^{4}. Let's rewrite the numerator:
s^{2} -4s +7 = s^{2} - 2s +1 -1 -2s +7 = (s-1)^{2} -2(s-1) + 4 (1)
If we divide (1) by (s-1) ^{4 }we will obtain:
1/(s-1)^{2} - 2 /(s-1) ^{3} + 4/(s-1)^{4 }(2)
But 1/(s-1) ^{n }is a Laplace trabnform of the function [1/(n-1)! t ^{ n-1} e^{t}.
If you consequitively apply this formula for n = 2, 3, 4 you will come up with function
y(t) = te^{t} - t^{2} e^{t} +(2/3) t^{3} e^{t}
Comments
Never mind. Can you just please tell me how to find the inverse Laplace transform of (s^2-4s+7)/(s-1)^4.
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