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The proportionality of $d^2y/dx^2$ and $y$
The 2019 Stack Overflow Developer Survey Results Are InIntegral $int_0^infty log(1+x^2)fraccoshfracpi x2sinh^2fracpi x2mathrm dx=2-frac4pi$Prove second derivative of $g$ is proportional to $g^2$Evaluating $int_0^2 fracdxsqrt[3]2x^2-x^3$Second order differential equation falling boxlaplace transform of smirnov density, i.e. how to calculate this integral?Showing the proportionality? (Calculus)Chemistry using integrationDifferential Equation.Proportionality, integrationcalculating $int_0 ^3 xsqrtdx$
$begingroup$
Suppose that $x$ and $y$ are related by the equation $displaystyle x=int_0^yfrac1sqrt1+4t^2,dt$. Show that $d^2y/dx^2$ is proportional to $y$ and find the constant of proportionality.
I try to calculate the problem and I got the proportionality is $4(1+4y^2)^-1/2$
But it is not a constant.
Thank you so much.
calculus
edited Mar 30 at 12:08
egreg
185k1486208
asked Mar 30 at 11:47
user630900user630900
12
$endgroup$
add a comment |
$begingroup$
Suppose that $x$ and $y$ are related by the equation $displaystyle x=int_0^yfrac1sqrt1+4t^2,dt$. Show that $d^2y/dx^2$ is proportional to $y$ and find the constant of proportionality.
I try to calculate the problem and I got the proportionality is $4(1+4y^2)^-1/2$
But it is not a constant.
Thank you so much.
calculus
edited Mar 30 at 12:08
egreg
185k1486208
asked Mar 30 at 11:47
user630900user630900
12
$endgroup$
$begingroup$
How do you get your result?
$endgroup$
– Oscar Lanzi
Mar 30 at 12:03
1
$begingroup$
I seen that I forgot the dy/dx at the end,thank you.
$endgroup$
– user630900
Mar 30 at 12:22
add a comment |
$begingroup$
Suppose that $x$ and $y$ are related by the equation $displaystyle x=int_0^yfrac1sqrt1+4t^2,dt$. Show that $d^2y/dx^2$ is proportional to $y$ and find the constant of proportionality.
I try to calculate the problem and I got the proportionality is $4(1+4y^2)^-1/2$
But it is not a constant.
Thank you so much.
calculus
edited Mar 30 at 12:08
egreg
185k1486208
asked Mar 30 at 11:47
user630900user630900
12
$endgroup$
Suppose that $x$ and $y$ are related by the equation $displaystyle x=int_0^yfrac1sqrt1+4t^2,dt$. Show that $d^2y/dx^2$ is proportional to $y$ and find the constant of proportionality.
I try to calculate the problem and I got the proportionality is $4(1+4y^2)^-1/2$
But it is not a constant.
Thank you so much.
calculus
calculus
edited Mar 30 at 12:08
egreg
185k1486208
asked Mar 30 at 11:47
user630900user630900
12
edited Mar 30 at 12:08
egreg
185k1486208
asked Mar 30 at 11:47
user630900user630900
12
edited Mar 30 at 12:08
egreg
185k1486208
edited Mar 30 at 12:08
egreg
185k1486208
edited Mar 30 at 12:08
egreg
185k1486208
185k1486208
asked Mar 30 at 11:47
user630900user630900
12
asked Mar 30 at 11:47
user630900user630900
12
asked Mar 30 at 11:47
user630900user630900
12
12
$begingroup$
How do you get your result?
$endgroup$
– Oscar Lanzi
Mar 30 at 12:03
1
$begingroup$
I seen that I forgot the dy/dx at the end,thank you.
$endgroup$
– user630900
Mar 30 at 12:22
add a comment |
$begingroup$
How do you get your result?
$endgroup$
– Oscar Lanzi
Mar 30 at 12:03
1
$begingroup$
I seen that I forgot the dy/dx at the end,thank you.
$endgroup$
– user630900
Mar 30 at 12:22
$begingroup$
How do you get your result?
$endgroup$
– Oscar Lanzi
Mar 30 at 12:03
$begingroup$
How do you get your result?
$endgroup$
– Oscar Lanzi
Mar 30 at 12:03
1
1
$begingroup$
I seen that I forgot the dy/dx at the end,thank you.
$endgroup$
– user630900
Mar 30 at 12:22
$begingroup$
I seen that I forgot the dy/dx at the end,thank you.
$endgroup$
– user630900
Mar 30 at 12:22
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Here is a solution that gives explicitly the expression of functions we are working on.
You may know that
$$textthe antiderivative of dfrac1sqrt1+t^2 textis x=sinh^-1(t)+k$$
where $k$ is an arbitrary constant (sinh$^-1$ being the inverse hyperbolic sine also denoted arcsinh).
Thus, using an elementary change of variable,
$$textthe antiderivative of dfrac1sqrt1+(2y)^2 is x=tfrac12 textsinh^-1(2y)+k tag1$$
As $x$ is expressed in terms of $y$, we are invited to reverse (1) for expressing $y$ as a function of $x$ ; doing this is easy : first, transform (1) into :
$$2(x-k)=textsinh^-1(2y)$$
then take the $sinh$ of LHS and RHS ; we get
$$sinh(2(x-k))=2y iff $$
$$y=tfrac12 sinh(2(x-k))tag2$$
From there, you can easily check that the second derivative of $y$ is $4$ times $y$.
Remark : Of course, the converse is not true. (2) is not, by far, the most general solution to the different and classical issue :
Find all functions proportional to their second derivative,
which has to be divided into two cases, according to the sign of this proportionality constant (of the form $k^2$ or $-k^2$) yielding two second order ordinary differential equations (ODE) :
$$fracd^2ydt^2=k^2 y textand fracd^2ydt^2=-k^2 y tag3$$
with resp. solutions :
$$y=A textsinh(kx+B) textand y=A sin(kx+B)$$
where $A$ and $B$ are arbitrary constants.
(The second ODE in (3) is called the ODE of harmonic oscillator).
answered Mar 30 at 16:06
Jean MarieJean Marie
31.5k42355
$endgroup$
$begingroup$
Thank you for your answer.
$endgroup$
– user630900
Mar 30 at 16:43
add a comment |
$begingroup$
$$fracdydx=left(fracdxdyright)^-1=left(fracddyint_0^yfrac1sqrt1+4t^2right)^-1=sqrt1+4y^2$$
so
$$fracd^2ydx^2=fracddxsqrt1+4y^2=frac4yfracdydxsqrt1+4y^2=4y$$
the proportionality is $4$.
answered Mar 30 at 12:07
Xin FuXin Fu
34319
$endgroup$
$begingroup$
Thank you very much.
$endgroup$
– user630900
Mar 30 at 12:23
add a comment |
$begingroup$
The constant is $4$: $frac dx dy =(1+4y^2)^-1/2$ so $frac dy dx =(1+4y^2)^1/2$. This gives $frac d^2y dx^2 =4y (1+4y^2)^-1/2frac dy dx=4y$.
answered Mar 30 at 12:12
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
$endgroup$
$begingroup$
Thank you very much.
$endgroup$
– user630900
Mar 30 at 12:23
add a comment |
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3 Answers
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3 Answers
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$begingroup$
Here is a solution that gives explicitly the expression of functions we are working on.
You may know that
$$textthe antiderivative of dfrac1sqrt1+t^2 textis x=sinh^-1(t)+k$$
where $k$ is an arbitrary constant (sinh$^-1$ being the inverse hyperbolic sine also denoted arcsinh).
Thus, using an elementary change of variable,
$$textthe antiderivative of dfrac1sqrt1+(2y)^2 is x=tfrac12 textsinh^-1(2y)+k tag1$$
As $x$ is expressed in terms of $y$, we are invited to reverse (1) for expressing $y$ as a function of $x$ ; doing this is easy : first, transform (1) into :
$$2(x-k)=textsinh^-1(2y)$$
then take the $sinh$ of LHS and RHS ; we get
$$sinh(2(x-k))=2y iff $$
$$y=tfrac12 sinh(2(x-k))tag2$$
From there, you can easily check that the second derivative of $y$ is $4$ times $y$.
Remark : Of course, the converse is not true. (2) is not, by far, the most general solution to the different and classical issue :
Find all functions proportional to their second derivative,
which has to be divided into two cases, according to the sign of this proportionality constant (of the form $k^2$ or $-k^2$) yielding two second order ordinary differential equations (ODE) :
$$fracd^2ydt^2=k^2 y textand fracd^2ydt^2=-k^2 y tag3$$
with resp. solutions :
$$y=A textsinh(kx+B) textand y=A sin(kx+B)$$
where $A$ and $B$ are arbitrary constants.
(The second ODE in (3) is called the ODE of harmonic oscillator).
answered Mar 30 at 16:06
Jean MarieJean Marie
31.5k42355
$endgroup$
$begingroup$
Thank you for your answer.
$endgroup$
– user630900
Mar 30 at 16:43
add a comment |
$begingroup$
Here is a solution that gives explicitly the expression of functions we are working on.
You may know that
$$textthe antiderivative of dfrac1sqrt1+t^2 textis x=sinh^-1(t)+k$$
where $k$ is an arbitrary constant (sinh$^-1$ being the inverse hyperbolic sine also denoted arcsinh).
Thus, using an elementary change of variable,
$$textthe antiderivative of dfrac1sqrt1+(2y)^2 is x=tfrac12 textsinh^-1(2y)+k tag1$$
As $x$ is expressed in terms of $y$, we are invited to reverse (1) for expressing $y$ as a function of $x$ ; doing this is easy : first, transform (1) into :
$$2(x-k)=textsinh^-1(2y)$$
then take the $sinh$ of LHS and RHS ; we get
$$sinh(2(x-k))=2y iff $$
$$y=tfrac12 sinh(2(x-k))tag2$$
From there, you can easily check that the second derivative of $y$ is $4$ times $y$.
Remark : Of course, the converse is not true. (2) is not, by far, the most general solution to the different and classical issue :
Find all functions proportional to their second derivative,
which has to be divided into two cases, according to the sign of this proportionality constant (of the form $k^2$ or $-k^2$) yielding two second order ordinary differential equations (ODE) :
$$fracd^2ydt^2=k^2 y textand fracd^2ydt^2=-k^2 y tag3$$
with resp. solutions :
$$y=A textsinh(kx+B) textand y=A sin(kx+B)$$
where $A$ and $B$ are arbitrary constants.
(The second ODE in (3) is called the ODE of harmonic oscillator).
answered Mar 30 at 16:06
Jean MarieJean Marie
31.5k42355
$endgroup$
$begingroup$
Thank you for your answer.
$endgroup$
– user630900
Mar 30 at 16:43
add a comment |
$begingroup$
Here is a solution that gives explicitly the expression of functions we are working on.
You may know that
$$textthe antiderivative of dfrac1sqrt1+t^2 textis x=sinh^-1(t)+k$$
where $k$ is an arbitrary constant (sinh$^-1$ being the inverse hyperbolic sine also denoted arcsinh).
Thus, using an elementary change of variable,
$$textthe antiderivative of dfrac1sqrt1+(2y)^2 is x=tfrac12 textsinh^-1(2y)+k tag1$$
As $x$ is expressed in terms of $y$, we are invited to reverse (1) for expressing $y$ as a function of $x$ ; doing this is easy : first, transform (1) into :
$$2(x-k)=textsinh^-1(2y)$$
then take the $sinh$ of LHS and RHS ; we get
$$sinh(2(x-k))=2y iff $$
$$y=tfrac12 sinh(2(x-k))tag2$$
From there, you can easily check that the second derivative of $y$ is $4$ times $y$.
Remark : Of course, the converse is not true. (2) is not, by far, the most general solution to the different and classical issue :
Find all functions proportional to their second derivative,
which has to be divided into two cases, according to the sign of this proportionality constant (of the form $k^2$ or $-k^2$) yielding two second order ordinary differential equations (ODE) :
$$fracd^2ydt^2=k^2 y textand fracd^2ydt^2=-k^2 y tag3$$
with resp. solutions :
$$y=A textsinh(kx+B) textand y=A sin(kx+B)$$
where $A$ and $B$ are arbitrary constants.
(The second ODE in (3) is called the ODE of harmonic oscillator).
answered Mar 30 at 16:06
Jean MarieJean Marie
31.5k42355
$endgroup$
Here is a solution that gives explicitly the expression of functions we are working on.
You may know that
$$textthe antiderivative of dfrac1sqrt1+t^2 textis x=sinh^-1(t)+k$$
where $k$ is an arbitrary constant (sinh$^-1$ being the inverse hyperbolic sine also denoted arcsinh).
Thus, using an elementary change of variable,
$$textthe antiderivative of dfrac1sqrt1+(2y)^2 is x=tfrac12 textsinh^-1(2y)+k tag1$$
As $x$ is expressed in terms of $y$, we are invited to reverse (1) for expressing $y$ as a function of $x$ ; doing this is easy : first, transform (1) into :
$$2(x-k)=textsinh^-1(2y)$$
then take the $sinh$ of LHS and RHS ; we get
$$sinh(2(x-k))=2y iff $$
$$y=tfrac12 sinh(2(x-k))tag2$$
From there, you can easily check that the second derivative of $y$ is $4$ times $y$.
Remark : Of course, the converse is not true. (2) is not, by far, the most general solution to the different and classical issue :
Find all functions proportional to their second derivative,
which has to be divided into two cases, according to the sign of this proportionality constant (of the form $k^2$ or $-k^2$) yielding two second order ordinary differential equations (ODE) :
$$fracd^2ydt^2=k^2 y textand fracd^2ydt^2=-k^2 y tag3$$
with resp. solutions :
$$y=A textsinh(kx+B) textand y=A sin(kx+B)$$
where $A$ and $B$ are arbitrary constants.
(The second ODE in (3) is called the ODE of harmonic oscillator).
answered Mar 30 at 16:06
Jean MarieJean Marie
31.5k42355
edited Mar 31 at 7:39
answered Mar 30 at 16:06
Jean MarieJean Marie
31.5k42355
answered Mar 30 at 16:06
Jean MarieJean Marie
31.5k42355
answered Mar 30 at 16:06
Jean MarieJean Marie
31.5k42355
31.5k42355
$begingroup$
Thank you for your answer.
$endgroup$
– user630900
Mar 30 at 16:43
add a comment |
$begingroup$
Thank you for your answer.
$endgroup$
– user630900
Mar 30 at 16:43
$begingroup$
Thank you for your answer.
$endgroup$
– user630900
Mar 30 at 16:43
$begingroup$
Thank you for your answer.
$endgroup$
– user630900
Mar 30 at 16:43
add a comment |
$begingroup$
$$fracdydx=left(fracdxdyright)^-1=left(fracddyint_0^yfrac1sqrt1+4t^2right)^-1=sqrt1+4y^2$$
so
$$fracd^2ydx^2=fracddxsqrt1+4y^2=frac4yfracdydxsqrt1+4y^2=4y$$
the proportionality is $4$.
answered Mar 30 at 12:07
Xin FuXin Fu
34319
$endgroup$
$begingroup$
Thank you very much.
$endgroup$
– user630900
Mar 30 at 12:23
add a comment |
$begingroup$
$$fracdydx=left(fracdxdyright)^-1=left(fracddyint_0^yfrac1sqrt1+4t^2right)^-1=sqrt1+4y^2$$
so
$$fracd^2ydx^2=fracddxsqrt1+4y^2=frac4yfracdydxsqrt1+4y^2=4y$$
the proportionality is $4$.
answered Mar 30 at 12:07
Xin FuXin Fu
34319
$endgroup$
$begingroup$
Thank you very much.
$endgroup$
– user630900
Mar 30 at 12:23
add a comment |
$begingroup$
$$fracdydx=left(fracdxdyright)^-1=left(fracddyint_0^yfrac1sqrt1+4t^2right)^-1=sqrt1+4y^2$$
so
$$fracd^2ydx^2=fracddxsqrt1+4y^2=frac4yfracdydxsqrt1+4y^2=4y$$
the proportionality is $4$.
answered Mar 30 at 12:07
Xin FuXin Fu
34319
$endgroup$
$$fracdydx=left(fracdxdyright)^-1=left(fracddyint_0^yfrac1sqrt1+4t^2right)^-1=sqrt1+4y^2$$
so
$$fracd^2ydx^2=fracddxsqrt1+4y^2=frac4yfracdydxsqrt1+4y^2=4y$$
the proportionality is $4$.
answered Mar 30 at 12:07
Xin FuXin Fu
34319
answered Mar 30 at 12:07
Xin FuXin Fu
34319
answered Mar 30 at 12:07
Xin FuXin Fu
34319
answered Mar 30 at 12:07
Xin FuXin Fu
34319
34319
$begingroup$
Thank you very much.
$endgroup$
– user630900
Mar 30 at 12:23
add a comment |
$begingroup$
Thank you very much.
$endgroup$
– user630900
Mar 30 at 12:23
$begingroup$
Thank you very much.
$endgroup$
– user630900
Mar 30 at 12:23
$begingroup$
Thank you very much.
$endgroup$
– user630900
Mar 30 at 12:23
add a comment |
$begingroup$
The constant is $4$: $frac dx dy =(1+4y^2)^-1/2$ so $frac dy dx =(1+4y^2)^1/2$. This gives $frac d^2y dx^2 =4y (1+4y^2)^-1/2frac dy dx=4y$.
answered Mar 30 at 12:12
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
$endgroup$
$begingroup$
Thank you very much.
$endgroup$
– user630900
Mar 30 at 12:23
add a comment |
$begingroup$
The constant is $4$: $frac dx dy =(1+4y^2)^-1/2$ so $frac dy dx =(1+4y^2)^1/2$. This gives $frac d^2y dx^2 =4y (1+4y^2)^-1/2frac dy dx=4y$.
answered Mar 30 at 12:12
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
$endgroup$
$begingroup$
Thank you very much.
$endgroup$
– user630900
Mar 30 at 12:23
add a comment |
$begingroup$
The constant is $4$: $frac dx dy =(1+4y^2)^-1/2$ so $frac dy dx =(1+4y^2)^1/2$. This gives $frac d^2y dx^2 =4y (1+4y^2)^-1/2frac dy dx=4y$.
answered Mar 30 at 12:12
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
$endgroup$
The constant is $4$: $frac dx dy =(1+4y^2)^-1/2$ so $frac dy dx =(1+4y^2)^1/2$. This gives $frac d^2y dx^2 =4y (1+4y^2)^-1/2frac dy dx=4y$.
answered Mar 30 at 12:12
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
answered Mar 30 at 12:12
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
answered Mar 30 at 12:12
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
answered Mar 30 at 12:12
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
73.6k53170
$begingroup$
Thank you very much.
$endgroup$
– user630900
Mar 30 at 12:23
add a comment |
$begingroup$
Thank you very much.
$endgroup$
– user630900
Mar 30 at 12:23
$begingroup$
Thank you very much.
$endgroup$
– user630900
Mar 30 at 12:23
$begingroup$
Thank you very much.
$endgroup$
– user630900
Mar 30 at 12:23
add a comment |
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$begingroup$
How do you get your result?
$endgroup$
– Oscar Lanzi
Mar 30 at 12:03
1
$begingroup$
I seen that I forgot the dy/dx at the end,thank you.
$endgroup$
– user630900
Mar 30 at 12:22