Mixing
How to find a specific curve if the initial value is not given?
$begingroup$
Question:
Let $y(x)$ be the solution of the differential equation
$xcdot ln(x)dfrac{dy}{dx}+y=2xcdot ln(x)$, $xge1$.
Find $y(e)$.
Answer: $y(e) = 2$
Problem:
So I understand that this can be converted into a simple linear differential equation and found that the solution is:
$ycdot ln(x)=2(xcdot ln(x) - x) + C$
This is a family of curves. However for solving the question, I need a specific curve out of all these.
What I don't understand is how how do I find that particular curve as the initial value of the function is not given.
calculus differential-equations
asked Dec 30 '18 at 11:20
harshit54harshit54
346113
$endgroup$
|
show 6 more comments
$begingroup$
Question:
Let $y(x)$ be the solution of the differential equation
$xcdot ln(x)dfrac{dy}{dx}+y=2xcdot ln(x)$, $xge1$.
Find $y(e)$.
Answer: $y(e) = 2$
Problem:
So I understand that this can be converted into a simple linear differential equation and found that the solution is:
$ycdot ln(x)=2(xcdot ln(x) - x) + C$
This is a family of curves. However for solving the question, I need a specific curve out of all these.
What I don't understand is how how do I find that particular curve as the initial value of the function is not given.
calculus differential-equations
asked Dec 30 '18 at 11:20
harshit54harshit54
346113
$endgroup$
$begingroup$
y(e) = C.......
$endgroup$
– William Elliot
Dec 30 '18 at 11:46
$begingroup$
@WilliamElliot Nah. Edited the question.
$endgroup$
– harshit54
Dec 30 '18 at 11:48
3
$begingroup$
Note that, evaluating the original equation with $x=1$, you find $y(1)=0$
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 11:59
$begingroup$
@MartínVacasVignolo So there is only one possible solution for this equation. But how is it possible if I am not given the initial values for the DE. Because the whole family satisfies the given DE.
$endgroup$
– harshit54
Dec 30 '18 at 12:06
1
$begingroup$
hmm no, without conditions you must consider the maximal domain. In this case $mathbb{R}^+$, and because $1inmathbb{R}^+$, is an equivalent problem
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 12:27
|
show 6 more comments
$begingroup$
Question:
Let $y(x)$ be the solution of the differential equation
$xcdot ln(x)dfrac{dy}{dx}+y=2xcdot ln(x)$, $xge1$.
Find $y(e)$.
Answer: $y(e) = 2$
Problem:
So I understand that this can be converted into a simple linear differential equation and found that the solution is:
$ycdot ln(x)=2(xcdot ln(x) - x) + C$
This is a family of curves. However for solving the question, I need a specific curve out of all these.
What I don't understand is how how do I find that particular curve as the initial value of the function is not given.
calculus differential-equations
asked Dec 30 '18 at 11:20
harshit54harshit54
346113
$endgroup$
Question:
Let $y(x)$ be the solution of the differential equation
$xcdot ln(x)dfrac{dy}{dx}+y=2xcdot ln(x)$, $xge1$.
Find $y(e)$.
Answer: $y(e) = 2$
Problem:
So I understand that this can be converted into a simple linear differential equation and found that the solution is:
$ycdot ln(x)=2(xcdot ln(x) - x) + C$
This is a family of curves. However for solving the question, I need a specific curve out of all these.
What I don't understand is how how do I find that particular curve as the initial value of the function is not given.
calculus differential-equations
calculus differential-equations
asked Dec 30 '18 at 11:20
harshit54harshit54
346113
asked Dec 30 '18 at 11:20
harshit54harshit54
346113
edited Dec 30 '18 at 11:47
harshit54
asked Dec 30 '18 at 11:20
harshit54harshit54
346113
asked Dec 30 '18 at 11:20
harshit54harshit54
346113
asked Dec 30 '18 at 11:20
harshit54harshit54
346113
346113
$begingroup$
y(e) = C.......
$endgroup$
– William Elliot
Dec 30 '18 at 11:46
$begingroup$
@WilliamElliot Nah. Edited the question.
$endgroup$
– harshit54
Dec 30 '18 at 11:48
3
$begingroup$
Note that, evaluating the original equation with $x=1$, you find $y(1)=0$
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 11:59
$begingroup$
@MartínVacasVignolo So there is only one possible solution for this equation. But how is it possible if I am not given the initial values for the DE. Because the whole family satisfies the given DE.
$endgroup$
– harshit54
Dec 30 '18 at 12:06
1
$begingroup$
hmm no, without conditions you must consider the maximal domain. In this case $mathbb{R}^+$, and because $1inmathbb{R}^+$, is an equivalent problem
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 12:27
|
show 6 more comments
$begingroup$
y(e) = C.......
$endgroup$
– William Elliot
Dec 30 '18 at 11:46
$begingroup$
@WilliamElliot Nah. Edited the question.
$endgroup$
– harshit54
Dec 30 '18 at 11:48
3
$begingroup$
Note that, evaluating the original equation with $x=1$, you find $y(1)=0$
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 11:59
$begingroup$
@MartínVacasVignolo So there is only one possible solution for this equation. But how is it possible if I am not given the initial values for the DE. Because the whole family satisfies the given DE.
$endgroup$
– harshit54
Dec 30 '18 at 12:06
1
$begingroup$
hmm no, without conditions you must consider the maximal domain. In this case $mathbb{R}^+$, and because $1inmathbb{R}^+$, is an equivalent problem
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 12:27
$begingroup$
y(e) = C.......
$endgroup$
– William Elliot
Dec 30 '18 at 11:46
$begingroup$
y(e) = C.......
$endgroup$
– William Elliot
Dec 30 '18 at 11:46
$begingroup$
@WilliamElliot Nah. Edited the question.
$endgroup$
– harshit54
Dec 30 '18 at 11:48
$begingroup$
@WilliamElliot Nah. Edited the question.
$endgroup$
– harshit54
Dec 30 '18 at 11:48
3
3
$begingroup$
Note that, evaluating the original equation with $x=1$, you find $y(1)=0$
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 11:59
$begingroup$
Note that, evaluating the original equation with $x=1$, you find $y(1)=0$
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 11:59
$begingroup$
@MartínVacasVignolo So there is only one possible solution for this equation. But how is it possible if I am not given the initial values for the DE. Because the whole family satisfies the given DE.
$endgroup$
– harshit54
Dec 30 '18 at 12:06
$begingroup$
@MartínVacasVignolo So there is only one possible solution for this equation. But how is it possible if I am not given the initial values for the DE. Because the whole family satisfies the given DE.
$endgroup$
– harshit54
Dec 30 '18 at 12:06
1
1
$begingroup$
hmm no, without conditions you must consider the maximal domain. In this case $mathbb{R}^+$, and because $1inmathbb{R}^+$, is an equivalent problem
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 12:27
$begingroup$
hmm no, without conditions you must consider the maximal domain. In this case $mathbb{R}^+$, and because $1inmathbb{R}^+$, is an equivalent problem
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 12:27
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I'd say that the answer $y(e) = 2$ is wrong if the question is stated like this.
You have already found the correct general solution of the first-order linear ODE, with one constant of integration $C in mathbb{R}$. From this you do indeed get $y(e) = C$ as William Elliot has pointed out. Therefore, you can obtain any value for $y(e)$, depending on the value of $C$.
The reason why the particular solution with $C=2$ is of interest is that it is the only solution of this ODE for which the (right-sided) limit as $x searrow 1$ is finite.
For the general solution $y(x) = frac{2x(ln(x)-1)+C}{ln(x)}$, $C in mathbb{R}$, we have
begin{equation}
lim_{x searrow 1} y(x) = left{ begin{array}{ll}
-infty, C < 2\
0, C = 2\
infty, C > 2
end{array}
right..
end{equation}
Therefore, if we assume that the right-sided limit of $y(x)$ is finite as $x$ approaches $1$, then there remains only the particular solution with $C=2$, which satisfies $y(e) = 2$.
But this assumption needs to be added to the question, otherwise the answer is wrong.
answered Jan 1 at 7:28
ChristophChristoph
1115
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
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$begingroup$
I'd say that the answer $y(e) = 2$ is wrong if the question is stated like this.
You have already found the correct general solution of the first-order linear ODE, with one constant of integration $C in mathbb{R}$. From this you do indeed get $y(e) = C$ as William Elliot has pointed out. Therefore, you can obtain any value for $y(e)$, depending on the value of $C$.
The reason why the particular solution with $C=2$ is of interest is that it is the only solution of this ODE for which the (right-sided) limit as $x searrow 1$ is finite.
For the general solution $y(x) = frac{2x(ln(x)-1)+C}{ln(x)}$, $C in mathbb{R}$, we have
begin{equation}
lim_{x searrow 1} y(x) = left{ begin{array}{ll}
-infty, C < 2\
0, C = 2\
infty, C > 2
end{array}
right..
end{equation}
Therefore, if we assume that the right-sided limit of $y(x)$ is finite as $x$ approaches $1$, then there remains only the particular solution with $C=2$, which satisfies $y(e) = 2$.
But this assumption needs to be added to the question, otherwise the answer is wrong.
answered Jan 1 at 7:28
ChristophChristoph
1115
$endgroup$
add a comment |
$begingroup$
I'd say that the answer $y(e) = 2$ is wrong if the question is stated like this.
You have already found the correct general solution of the first-order linear ODE, with one constant of integration $C in mathbb{R}$. From this you do indeed get $y(e) = C$ as William Elliot has pointed out. Therefore, you can obtain any value for $y(e)$, depending on the value of $C$.
The reason why the particular solution with $C=2$ is of interest is that it is the only solution of this ODE for which the (right-sided) limit as $x searrow 1$ is finite.
For the general solution $y(x) = frac{2x(ln(x)-1)+C}{ln(x)}$, $C in mathbb{R}$, we have
begin{equation}
lim_{x searrow 1} y(x) = left{ begin{array}{ll}
-infty, C < 2\
0, C = 2\
infty, C > 2
end{array}
right..
end{equation}
Therefore, if we assume that the right-sided limit of $y(x)$ is finite as $x$ approaches $1$, then there remains only the particular solution with $C=2$, which satisfies $y(e) = 2$.
But this assumption needs to be added to the question, otherwise the answer is wrong.
answered Jan 1 at 7:28
ChristophChristoph
1115
$endgroup$
add a comment |
$begingroup$
I'd say that the answer $y(e) = 2$ is wrong if the question is stated like this.
You have already found the correct general solution of the first-order linear ODE, with one constant of integration $C in mathbb{R}$. From this you do indeed get $y(e) = C$ as William Elliot has pointed out. Therefore, you can obtain any value for $y(e)$, depending on the value of $C$.
The reason why the particular solution with $C=2$ is of interest is that it is the only solution of this ODE for which the (right-sided) limit as $x searrow 1$ is finite.
For the general solution $y(x) = frac{2x(ln(x)-1)+C}{ln(x)}$, $C in mathbb{R}$, we have
begin{equation}
lim_{x searrow 1} y(x) = left{ begin{array}{ll}
-infty, C < 2\
0, C = 2\
infty, C > 2
end{array}
right..
end{equation}
Therefore, if we assume that the right-sided limit of $y(x)$ is finite as $x$ approaches $1$, then there remains only the particular solution with $C=2$, which satisfies $y(e) = 2$.
But this assumption needs to be added to the question, otherwise the answer is wrong.
answered Jan 1 at 7:28
ChristophChristoph
1115
$endgroup$
I'd say that the answer $y(e) = 2$ is wrong if the question is stated like this.
You have already found the correct general solution of the first-order linear ODE, with one constant of integration $C in mathbb{R}$. From this you do indeed get $y(e) = C$ as William Elliot has pointed out. Therefore, you can obtain any value for $y(e)$, depending on the value of $C$.
The reason why the particular solution with $C=2$ is of interest is that it is the only solution of this ODE for which the (right-sided) limit as $x searrow 1$ is finite.
For the general solution $y(x) = frac{2x(ln(x)-1)+C}{ln(x)}$, $C in mathbb{R}$, we have
begin{equation}
lim_{x searrow 1} y(x) = left{ begin{array}{ll}
-infty, C < 2\
0, C = 2\
infty, C > 2
end{array}
right..
end{equation}
Therefore, if we assume that the right-sided limit of $y(x)$ is finite as $x$ approaches $1$, then there remains only the particular solution with $C=2$, which satisfies $y(e) = 2$.
But this assumption needs to be added to the question, otherwise the answer is wrong.
answered Jan 1 at 7:28
ChristophChristoph
1115
answered Jan 1 at 7:28
ChristophChristoph
1115
answered Jan 1 at 7:28
ChristophChristoph
1115
answered Jan 1 at 7:28
ChristophChristoph
1115
1115
add a comment |
add a comment |
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$begingroup$
y(e) = C.......
$endgroup$
– William Elliot
Dec 30 '18 at 11:46
$begingroup$
@WilliamElliot Nah. Edited the question.
$endgroup$
– harshit54
Dec 30 '18 at 11:48
3
$begingroup$
Note that, evaluating the original equation with $x=1$, you find $y(1)=0$
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 11:59
$begingroup$
@MartínVacasVignolo So there is only one possible solution for this equation. But how is it possible if I am not given the initial values for the DE. Because the whole family satisfies the given DE.
$endgroup$
– harshit54
Dec 30 '18 at 12:06
1
$begingroup$
hmm no, without conditions you must consider the maximal domain. In this case $mathbb{R}^+$, and because $1inmathbb{R}^+$, is an equivalent problem
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 12:27