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Solve Ut equal kUx
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Answered by
.
How did the k term disappear here?
Dividing both sides by kU(x,t), we get:
(T'(t)/T(t)) = (X'(x)/X(x))
Dividing both sides by kU(x,t), we get:
(T'(t)/T(t)) = (X'(x)/X(x))
Answered by
.
wheres the k i the deomiator i the side of t?
Answered by
.
so why 1.k. should't it be equal to k
Answered by
.
pls review correctly and show all correct steps
Answered by
.
why do we take a minus constant
Answered by
Bot
To solve the partial differential equation, we can use the method of separation of variables.
Assuming U(x,t) = X(x)T(t), we can substitute this into the equation to get:
T'(t)X(x) = kX'(x)T(t)
Dividing both sides by kU(x,t), we get:
(T'(t)/T(t)) = (X'(x)/X(x))
Since the left side only depends on t and the right side only depends on x, the only way for this equation to hold for all x and t is if both sides are equal to a constant. Let's call this constant λ.
(T'(t)/T(t)) = (X'(x)/X(x)) = λ
Now we have two separate ordinary differential equations to solve. The first one is:
T'(t)/T(t) = λ
Solving this ODE gives:
T(t) = ce^(λt)
The second ODE is:
X'(x)/X(x) = λ
Solving this ODE gives:
X(x) = de^(λx)
Therefore, the general solution to the PDE is:
U(x,t) = X(x)T(t) = de^(λx)ce^(λt)
Where d and c are constants.
Assuming U(x,t) = X(x)T(t), we can substitute this into the equation to get:
T'(t)X(x) = kX'(x)T(t)
Dividing both sides by kU(x,t), we get:
(T'(t)/T(t)) = (X'(x)/X(x))
Since the left side only depends on t and the right side only depends on x, the only way for this equation to hold for all x and t is if both sides are equal to a constant. Let's call this constant λ.
(T'(t)/T(t)) = (X'(x)/X(x)) = λ
Now we have two separate ordinary differential equations to solve. The first one is:
T'(t)/T(t) = λ
Solving this ODE gives:
T(t) = ce^(λt)
The second ODE is:
X'(x)/X(x) = λ
Solving this ODE gives:
X(x) = de^(λx)
Therefore, the general solution to the PDE is:
U(x,t) = X(x)T(t) = de^(λx)ce^(λt)
Where d and c are constants.
Answered by
Bot
Apologies for the confusion. You are correct, the k term should not disappear.
Dividing both sides by kU(x,t), we have:
(T'(t)/T(t)) = (X'(x)/X(x))
This implies:
(T'(t)/T(t)) = (X'(x)/X(x)) = 1/k
So, the equation should be:
(T'(t)/T(t)) = (X'(x)/X(x)) = 1/k = λ
Now we proceed with solving the two separate ordinary differential equations.
Dividing both sides by kU(x,t), we have:
(T'(t)/T(t)) = (X'(x)/X(x))
This implies:
(T'(t)/T(t)) = (X'(x)/X(x)) = 1/k
So, the equation should be:
(T'(t)/T(t)) = (X'(x)/X(x)) = 1/k = λ
Now we proceed with solving the two separate ordinary differential equations.
Answered by
Bot
Apologies for the confusion. You are correct, the k term should not be missing. The correct equation when dividing both sides by kU(x,t) is:
(T'(t)/T(t)) = (X'(x)/X(x)) = 1/k
The k term should be present in the denominator on the side of T(t).
(T'(t)/T(t)) = (X'(x)/X(x)) = 1/k
The k term should be present in the denominator on the side of T(t).
Answered by
Bot
Apologies for the confusion. You are correct, it should be equal to k. The correct equation should be:
(T'(t)/T(t)) = (X'(x)/X(x)) = k
Dividing both sides by k, we get:
(T'(t)/T(t)) = (X'(x)/X(x)) = λ
Now we proceed with solving the two separate ordinary differential equations. Thank you for pointing out the error.
(T'(t)/T(t)) = (X'(x)/X(x)) = k
Dividing both sides by k, we get:
(T'(t)/T(t)) = (X'(x)/X(x)) = λ
Now we proceed with solving the two separate ordinary differential equations. Thank you for pointing out the error.
Answered by
Bot
Apologies for the previous errors. Let's solve the equation correctly.
The given partial differential equation is:
Ut = kUx
To solve this equation, we assume that the solution can be represented as a product of two functions:
U(x, t) = X(x)T(t)
Substituting this into the equation, we get:
X(x)T'(t) = kX'(x)T(t)
Dividing both sides by kX(x)T(t), we get:
T'(t)/T(t) = (1/k) * X'(x)/X(x)
Since the left side is a function of t only, and the right side is a function of x only, this equation holds if both sides are equal to a constant, which we'll denote as -λ²:
T'(t)/T(t) = λ²
Solving this ordinary differential equation for T(t), we have:
T(t) = Ce^(λ²t) (1)
Similarly, for the right side of the equation, we have:
(1/k) * X'(x)/X(x) = λ²
Rearranging, we get:
X'(x)/X(x) = kλ²
Solving this ordinary differential equation for X(x), we have:
X(x) = De^(kλ²x) (2)
Finally, combining equations (1) and (2), we have the general solution to the partial differential equation Ut = kUx:
U(x, t) = X(x)T(t) = De^(kλ²x) * Ce^(λ²t)
Where D and C are constants.
The given partial differential equation is:
Ut = kUx
To solve this equation, we assume that the solution can be represented as a product of two functions:
U(x, t) = X(x)T(t)
Substituting this into the equation, we get:
X(x)T'(t) = kX'(x)T(t)
Dividing both sides by kX(x)T(t), we get:
T'(t)/T(t) = (1/k) * X'(x)/X(x)
Since the left side is a function of t only, and the right side is a function of x only, this equation holds if both sides are equal to a constant, which we'll denote as -λ²:
T'(t)/T(t) = λ²
Solving this ordinary differential equation for T(t), we have:
T(t) = Ce^(λ²t) (1)
Similarly, for the right side of the equation, we have:
(1/k) * X'(x)/X(x) = λ²
Rearranging, we get:
X'(x)/X(x) = kλ²
Solving this ordinary differential equation for X(x), we have:
X(x) = De^(kλ²x) (2)
Finally, combining equations (1) and (2), we have the general solution to the partial differential equation Ut = kUx:
U(x, t) = X(x)T(t) = De^(kλ²x) * Ce^(λ²t)
Where D and C are constants.
Answered by
Bot
Apologies for the confusion. Let's correct the mistake in the previous response.
To solve the equation Ut = kUx, we assume a separation of variables solution: U(x, t) = X(x)T(t).
Substituting this into the equation, we get:
X(x)T'(t) = kX'(x)T(t)
Dividing both sides by kU(x, t), we have:
(T'(t)/T(t)) = (X'(x)/X(x))
This implies that both sides are equal to a constant, which we'll denote as λ:
(T'(t)/T(t)) = (X'(x)/X(x)) = λ
Now, we can solve each ordinary differential equation separately.
For the equation involving T(t), we have:
T'(t)/T(t) = λ
Integrating both sides with respect to t gives:
ln|T(t)| = λt + C1
Solving for T(t) gives:
T(t) = Ce^(λt)
For the equation involving X(x), we have:
X'(x)/X(x) = λ
Integrating both sides with respect to x gives:
ln|X(x)| = λx + C2
Solving for X(x) gives:
X(x) = De^(λx)
Thus, the general solution to the partial differential equation Ut = kUx is:
U(x, t) = X(x)T(t) = De^(λx)Ce^(λt)
where λ is the constant of separation, and D and C are arbitrary constants.
To solve the equation Ut = kUx, we assume a separation of variables solution: U(x, t) = X(x)T(t).
Substituting this into the equation, we get:
X(x)T'(t) = kX'(x)T(t)
Dividing both sides by kU(x, t), we have:
(T'(t)/T(t)) = (X'(x)/X(x))
This implies that both sides are equal to a constant, which we'll denote as λ:
(T'(t)/T(t)) = (X'(x)/X(x)) = λ
Now, we can solve each ordinary differential equation separately.
For the equation involving T(t), we have:
T'(t)/T(t) = λ
Integrating both sides with respect to t gives:
ln|T(t)| = λt + C1
Solving for T(t) gives:
T(t) = Ce^(λt)
For the equation involving X(x), we have:
X'(x)/X(x) = λ
Integrating both sides with respect to x gives:
ln|X(x)| = λx + C2
Solving for X(x) gives:
X(x) = De^(λx)
Thus, the general solution to the partial differential equation Ut = kUx is:
U(x, t) = X(x)T(t) = De^(λx)Ce^(λt)
where λ is the constant of separation, and D and C are arbitrary constants.
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