Mixing
A second order differential equations with initial conditions solved using Laplace Transforms
$begingroup$
Below is a problem I did from the book Differential Equations by K.A. Stroud and Dexter Booth. I got the right answer but I am not sure I did it right especially when I took the inverse Laplace Transform. Therefore, I am hoping that somebody here can check it for me.
Thanks,
Bob
Problem:
Solve the following differential equation:
$$ f''(x) - 5f'(x) + 6f(x) = e^{2x} text{ where } f(0) = 0 text{ and } f'(0) = 0 $$
newline
Answer:
To solve this equation, I am going to use the Laplace transform.
begin{align*}
s^2F(s) - sf(0) - f'(0) - 5(sF(s) - f(0)) + 6F(s) &= frac{1}{s-2} \
s^2F(s) - 5(sF(s)) + 6F(s) &= frac{1}{s-2} \
(s^2-5s+6)F(s) &= frac{1}{s-2} \
F(s) &= frac{1}{(s-2)^2(s-3)} \
end{align*}
Now we apply the technique of partial fractions to find $f(x)$.
begin{align*}
frac{1}{(s-2)^2(s-3)} &= frac{A}{s-2} + frac{B}{(s-2)^2} + frac{C}{s-3} \
1 &= A(s-2)(s-3) + B(s-3) + C(s-2)^2 \
end{align*}
At $s = 2$ we have: $1 = B(2-3) = -B$ or $B = -1$. At $s =3$ we haveL $1 = C(3-2)^2 = C = 1$.
begin{align*}
A(s-2)(s-3) - (s-3) + (s-2)^2 &= 1 \
A + 1 &= 0 \
A &= -1 \
end{align*}
begin{align*}
F(s) &= frac{-1}{s-2} - frac{1}{(s-2)^2} + frac{1}{s-3} \
mathcal{L}^{-1}left( frac{1}{s-2} right) &= e^{-2x} \
end{align*}
Now, we need to find the inverse Laplace transform of:
$$ frac{1}{(s-2)^2} $$
Recall that if $mathcal{L}(f(t)) = F(s)$ then $mathcal{L}(e^{at}f(t)) = F(s-a)$
In this case, we use $a = 2$, $f(t) = x$ and $F(s) = frac{1}{s^2}$.
begin{align*}
mathcal{L}^{-1}left( frac{1}{(s-2)^2} right) &= xe^{2x} \
mathcal{L}^{-1}left( frac{1}{s-3} right) &= e^{3x} \
f(x) &= -e^{-2x} - xe^{-2x} + e^{3x} \
f(x) &= e^{3x} - (x+1 )e^{2x} \
end{align*}
ordinary-differential-equations laplace-transform
asked Jan 19 at 23:57
BobBob
923515
$endgroup$
add a comment |
$begingroup$
Below is a problem I did from the book Differential Equations by K.A. Stroud and Dexter Booth. I got the right answer but I am not sure I did it right especially when I took the inverse Laplace Transform. Therefore, I am hoping that somebody here can check it for me.
Thanks,
Bob
Problem:
Solve the following differential equation:
$$ f''(x) - 5f'(x) + 6f(x) = e^{2x} text{ where } f(0) = 0 text{ and } f'(0) = 0 $$
newline
Answer:
To solve this equation, I am going to use the Laplace transform.
begin{align*}
s^2F(s) - sf(0) - f'(0) - 5(sF(s) - f(0)) + 6F(s) &= frac{1}{s-2} \
s^2F(s) - 5(sF(s)) + 6F(s) &= frac{1}{s-2} \
(s^2-5s+6)F(s) &= frac{1}{s-2} \
F(s) &= frac{1}{(s-2)^2(s-3)} \
end{align*}
Now we apply the technique of partial fractions to find $f(x)$.
begin{align*}
frac{1}{(s-2)^2(s-3)} &= frac{A}{s-2} + frac{B}{(s-2)^2} + frac{C}{s-3} \
1 &= A(s-2)(s-3) + B(s-3) + C(s-2)^2 \
end{align*}
At $s = 2$ we have: $1 = B(2-3) = -B$ or $B = -1$. At $s =3$ we haveL $1 = C(3-2)^2 = C = 1$.
begin{align*}
A(s-2)(s-3) - (s-3) + (s-2)^2 &= 1 \
A + 1 &= 0 \
A &= -1 \
end{align*}
begin{align*}
F(s) &= frac{-1}{s-2} - frac{1}{(s-2)^2} + frac{1}{s-3} \
mathcal{L}^{-1}left( frac{1}{s-2} right) &= e^{-2x} \
end{align*}
Now, we need to find the inverse Laplace transform of:
$$ frac{1}{(s-2)^2} $$
Recall that if $mathcal{L}(f(t)) = F(s)$ then $mathcal{L}(e^{at}f(t)) = F(s-a)$
In this case, we use $a = 2$, $f(t) = x$ and $F(s) = frac{1}{s^2}$.
begin{align*}
mathcal{L}^{-1}left( frac{1}{(s-2)^2} right) &= xe^{2x} \
mathcal{L}^{-1}left( frac{1}{s-3} right) &= e^{3x} \
f(x) &= -e^{-2x} - xe^{-2x} + e^{3x} \
f(x) &= e^{3x} - (x+1 )e^{2x} \
end{align*}
ordinary-differential-equations laplace-transform
asked Jan 19 at 23:57
BobBob
923515
$endgroup$
add a comment |
$begingroup$
Below is a problem I did from the book Differential Equations by K.A. Stroud and Dexter Booth. I got the right answer but I am not sure I did it right especially when I took the inverse Laplace Transform. Therefore, I am hoping that somebody here can check it for me.
Thanks,
Bob
Problem:
Solve the following differential equation:
$$ f''(x) - 5f'(x) + 6f(x) = e^{2x} text{ where } f(0) = 0 text{ and } f'(0) = 0 $$
newline
Answer:
To solve this equation, I am going to use the Laplace transform.
begin{align*}
s^2F(s) - sf(0) - f'(0) - 5(sF(s) - f(0)) + 6F(s) &= frac{1}{s-2} \
s^2F(s) - 5(sF(s)) + 6F(s) &= frac{1}{s-2} \
(s^2-5s+6)F(s) &= frac{1}{s-2} \
F(s) &= frac{1}{(s-2)^2(s-3)} \
end{align*}
Now we apply the technique of partial fractions to find $f(x)$.
begin{align*}
frac{1}{(s-2)^2(s-3)} &= frac{A}{s-2} + frac{B}{(s-2)^2} + frac{C}{s-3} \
1 &= A(s-2)(s-3) + B(s-3) + C(s-2)^2 \
end{align*}
At $s = 2$ we have: $1 = B(2-3) = -B$ or $B = -1$. At $s =3$ we haveL $1 = C(3-2)^2 = C = 1$.
begin{align*}
A(s-2)(s-3) - (s-3) + (s-2)^2 &= 1 \
A + 1 &= 0 \
A &= -1 \
end{align*}
begin{align*}
F(s) &= frac{-1}{s-2} - frac{1}{(s-2)^2} + frac{1}{s-3} \
mathcal{L}^{-1}left( frac{1}{s-2} right) &= e^{-2x} \
end{align*}
Now, we need to find the inverse Laplace transform of:
$$ frac{1}{(s-2)^2} $$
Recall that if $mathcal{L}(f(t)) = F(s)$ then $mathcal{L}(e^{at}f(t)) = F(s-a)$
In this case, we use $a = 2$, $f(t) = x$ and $F(s) = frac{1}{s^2}$.
begin{align*}
mathcal{L}^{-1}left( frac{1}{(s-2)^2} right) &= xe^{2x} \
mathcal{L}^{-1}left( frac{1}{s-3} right) &= e^{3x} \
f(x) &= -e^{-2x} - xe^{-2x} + e^{3x} \
f(x) &= e^{3x} - (x+1 )e^{2x} \
end{align*}
ordinary-differential-equations laplace-transform
asked Jan 19 at 23:57
BobBob
923515
$endgroup$
Below is a problem I did from the book Differential Equations by K.A. Stroud and Dexter Booth. I got the right answer but I am not sure I did it right especially when I took the inverse Laplace Transform. Therefore, I am hoping that somebody here can check it for me.
Thanks,
Bob
Problem:
Solve the following differential equation:
$$ f''(x) - 5f'(x) + 6f(x) = e^{2x} text{ where } f(0) = 0 text{ and } f'(0) = 0 $$
newline
Answer:
To solve this equation, I am going to use the Laplace transform.
begin{align*}
s^2F(s) - sf(0) - f'(0) - 5(sF(s) - f(0)) + 6F(s) &= frac{1}{s-2} \
s^2F(s) - 5(sF(s)) + 6F(s) &= frac{1}{s-2} \
(s^2-5s+6)F(s) &= frac{1}{s-2} \
F(s) &= frac{1}{(s-2)^2(s-3)} \
end{align*}
Now we apply the technique of partial fractions to find $f(x)$.
begin{align*}
frac{1}{(s-2)^2(s-3)} &= frac{A}{s-2} + frac{B}{(s-2)^2} + frac{C}{s-3} \
1 &= A(s-2)(s-3) + B(s-3) + C(s-2)^2 \
end{align*}
At $s = 2$ we have: $1 = B(2-3) = -B$ or $B = -1$. At $s =3$ we haveL $1 = C(3-2)^2 = C = 1$.
begin{align*}
A(s-2)(s-3) - (s-3) + (s-2)^2 &= 1 \
A + 1 &= 0 \
A &= -1 \
end{align*}
begin{align*}
F(s) &= frac{-1}{s-2} - frac{1}{(s-2)^2} + frac{1}{s-3} \
mathcal{L}^{-1}left( frac{1}{s-2} right) &= e^{-2x} \
end{align*}
Now, we need to find the inverse Laplace transform of:
$$ frac{1}{(s-2)^2} $$
Recall that if $mathcal{L}(f(t)) = F(s)$ then $mathcal{L}(e^{at}f(t)) = F(s-a)$
In this case, we use $a = 2$, $f(t) = x$ and $F(s) = frac{1}{s^2}$.
begin{align*}
mathcal{L}^{-1}left( frac{1}{(s-2)^2} right) &= xe^{2x} \
mathcal{L}^{-1}left( frac{1}{s-3} right) &= e^{3x} \
f(x) &= -e^{-2x} - xe^{-2x} + e^{3x} \
f(x) &= e^{3x} - (x+1 )e^{2x} \
end{align*}
ordinary-differential-equations laplace-transform
ordinary-differential-equations laplace-transform
asked Jan 19 at 23:57
BobBob
923515
asked Jan 19 at 23:57
BobBob
923515
edited Jan 20 at 1:25
Bob
asked Jan 19 at 23:57
BobBob
923515
asked Jan 19 at 23:57
BobBob
923515
asked Jan 19 at 23:57
BobBob
923515
923515
add a comment |
add a comment |
1 Answer
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$begingroup$
Please check your work and make sure that the exponents are typed corrrectly. $$begin{align*}
mathcal{L}^{-1}left( frac{1}{(s-2)^2} right) &= xe^{2x} \
mathcal{L}^{-1}left( frac{1}{s-3} right) &= e^{3x} \
f(x) &= -e^{-2x} - xe^{-2x} + e^{3x} \
f(x) &= e^{3x} - (x+1 )e^{2x} \
end{align*}$$
While your first and second line above are correct, the third line has negative numbers in the exponents.
Over all your work is correct and once you take care of the typos, everything should be fine.
answered Jan 20 at 1:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
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add a comment |
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1 Answer
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$begingroup$
Please check your work and make sure that the exponents are typed corrrectly. $$begin{align*}
mathcal{L}^{-1}left( frac{1}{(s-2)^2} right) &= xe^{2x} \
mathcal{L}^{-1}left( frac{1}{s-3} right) &= e^{3x} \
f(x) &= -e^{-2x} - xe^{-2x} + e^{3x} \
f(x) &= e^{3x} - (x+1 )e^{2x} \
end{align*}$$
While your first and second line above are correct, the third line has negative numbers in the exponents.
Over all your work is correct and once you take care of the typos, everything should be fine.
answered Jan 20 at 1:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
Please check your work and make sure that the exponents are typed corrrectly. $$begin{align*}
mathcal{L}^{-1}left( frac{1}{(s-2)^2} right) &= xe^{2x} \
mathcal{L}^{-1}left( frac{1}{s-3} right) &= e^{3x} \
f(x) &= -e^{-2x} - xe^{-2x} + e^{3x} \
f(x) &= e^{3x} - (x+1 )e^{2x} \
end{align*}$$
While your first and second line above are correct, the third line has negative numbers in the exponents.
Over all your work is correct and once you take care of the typos, everything should be fine.
answered Jan 20 at 1:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
Please check your work and make sure that the exponents are typed corrrectly. $$begin{align*}
mathcal{L}^{-1}left( frac{1}{(s-2)^2} right) &= xe^{2x} \
mathcal{L}^{-1}left( frac{1}{s-3} right) &= e^{3x} \
f(x) &= -e^{-2x} - xe^{-2x} + e^{3x} \
f(x) &= e^{3x} - (x+1 )e^{2x} \
end{align*}$$
While your first and second line above are correct, the third line has negative numbers in the exponents.
Over all your work is correct and once you take care of the typos, everything should be fine.
answered Jan 20 at 1:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
Please check your work and make sure that the exponents are typed corrrectly. $$begin{align*}
mathcal{L}^{-1}left( frac{1}{(s-2)^2} right) &= xe^{2x} \
mathcal{L}^{-1}left( frac{1}{s-3} right) &= e^{3x} \
f(x) &= -e^{-2x} - xe^{-2x} + e^{3x} \
f(x) &= e^{3x} - (x+1 )e^{2x} \
end{align*}$$
While your first and second line above are correct, the third line has negative numbers in the exponents.
Over all your work is correct and once you take care of the typos, everything should be fine.
answered Jan 20 at 1:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 20 at 1:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 20 at 1:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 20 at 1:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
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