Mixing
Finding a differential equations given a general solution
$begingroup$
Let $A,B in mathbb{R}$. Find a differential equation having the
general solution $$ y(x) = A (x+ B)^n $$
ATTEMPT
Note that $y' = n A (x + B)^{n-1} $. Also, $y'' = n(n-1)A (x+B)^{n-2}$
Note that $(n-1)(x+B)^{-1} y' = y'' $. Thus, a differential equation can be
$$ frac{(n-1)}{x+B}y' - y'' = 0$$
or
$$ (n-1) y' - (x+B) y'' = 0$$
seemts like a trivial problem, but perhaps Im misunderstanding it?
am I correct?
ordinary-differential-equations
asked Jan 5 at 22:53
Jimmy SabaterJimmy Sabater
2,267319
$endgroup$
add a comment |
$begingroup$
Let $A,B in mathbb{R}$. Find a differential equation having the
general solution $$ y(x) = A (x+ B)^n $$
ATTEMPT
Note that $y' = n A (x + B)^{n-1} $. Also, $y'' = n(n-1)A (x+B)^{n-2}$
Note that $(n-1)(x+B)^{-1} y' = y'' $. Thus, a differential equation can be
$$ frac{(n-1)}{x+B}y' - y'' = 0$$
or
$$ (n-1) y' - (x+B) y'' = 0$$
seemts like a trivial problem, but perhaps Im misunderstanding it?
am I correct?
ordinary-differential-equations
asked Jan 5 at 22:53
Jimmy SabaterJimmy Sabater
2,267319
$endgroup$
$begingroup$
Your equation is wrong, check the degrees of all terms...
$endgroup$
– Mindlack
Jan 5 at 22:55
$begingroup$
I fixed it now.
$endgroup$
– Jimmy Sabater
Jan 5 at 22:55
$begingroup$
This is a very "soft" question, there is more than one answer. But the problem is that your answer keeps $B$ in it, which should be some sort of integration constant, not a known parameter of the differential equation.
$endgroup$
– orion
Jan 5 at 23:00
$begingroup$
should I expand the binomial ?
$endgroup$
– Jimmy Sabater
Jan 5 at 23:01
$begingroup$
Your differential equation indeed has $A(x+B)^n$ as a solution. It is not, however, the only one, but I do not know whether this is relevant.
$endgroup$
– Mindlack
Jan 5 at 23:03
add a comment |
$begingroup$
Let $A,B in mathbb{R}$. Find a differential equation having the
general solution $$ y(x) = A (x+ B)^n $$
ATTEMPT
Note that $y' = n A (x + B)^{n-1} $. Also, $y'' = n(n-1)A (x+B)^{n-2}$
Note that $(n-1)(x+B)^{-1} y' = y'' $. Thus, a differential equation can be
$$ frac{(n-1)}{x+B}y' - y'' = 0$$
or
$$ (n-1) y' - (x+B) y'' = 0$$
seemts like a trivial problem, but perhaps Im misunderstanding it?
am I correct?
ordinary-differential-equations
asked Jan 5 at 22:53
Jimmy SabaterJimmy Sabater
2,267319
$endgroup$
Let $A,B in mathbb{R}$. Find a differential equation having the
general solution $$ y(x) = A (x+ B)^n $$
ATTEMPT
Note that $y' = n A (x + B)^{n-1} $. Also, $y'' = n(n-1)A (x+B)^{n-2}$
Note that $(n-1)(x+B)^{-1} y' = y'' $. Thus, a differential equation can be
$$ frac{(n-1)}{x+B}y' - y'' = 0$$
or
$$ (n-1) y' - (x+B) y'' = 0$$
seemts like a trivial problem, but perhaps Im misunderstanding it?
am I correct?
ordinary-differential-equations
ordinary-differential-equations
asked Jan 5 at 22:53
Jimmy SabaterJimmy Sabater
2,267319
asked Jan 5 at 22:53
Jimmy SabaterJimmy Sabater
2,267319
edited Jan 5 at 22:55
Jimmy Sabater
asked Jan 5 at 22:53
Jimmy SabaterJimmy Sabater
2,267319
asked Jan 5 at 22:53
Jimmy SabaterJimmy Sabater
2,267319
asked Jan 5 at 22:53
Jimmy SabaterJimmy Sabater
2,267319
2,267319
$begingroup$
Your equation is wrong, check the degrees of all terms...
$endgroup$
– Mindlack
Jan 5 at 22:55
$begingroup$
I fixed it now.
$endgroup$
– Jimmy Sabater
Jan 5 at 22:55
$begingroup$
This is a very "soft" question, there is more than one answer. But the problem is that your answer keeps $B$ in it, which should be some sort of integration constant, not a known parameter of the differential equation.
$endgroup$
– orion
Jan 5 at 23:00
$begingroup$
should I expand the binomial ?
$endgroup$
– Jimmy Sabater
Jan 5 at 23:01
$begingroup$
Your differential equation indeed has $A(x+B)^n$ as a solution. It is not, however, the only one, but I do not know whether this is relevant.
$endgroup$
– Mindlack
Jan 5 at 23:03
add a comment |
$begingroup$
Your equation is wrong, check the degrees of all terms...
$endgroup$
– Mindlack
Jan 5 at 22:55
$begingroup$
I fixed it now.
$endgroup$
– Jimmy Sabater
Jan 5 at 22:55
$begingroup$
This is a very "soft" question, there is more than one answer. But the problem is that your answer keeps $B$ in it, which should be some sort of integration constant, not a known parameter of the differential equation.
$endgroup$
– orion
Jan 5 at 23:00
$begingroup$
should I expand the binomial ?
$endgroup$
– Jimmy Sabater
Jan 5 at 23:01
$begingroup$
Your differential equation indeed has $A(x+B)^n$ as a solution. It is not, however, the only one, but I do not know whether this is relevant.
$endgroup$
– Mindlack
Jan 5 at 23:03
$begingroup$
Your equation is wrong, check the degrees of all terms...
$endgroup$
– Mindlack
Jan 5 at 22:55
$begingroup$
Your equation is wrong, check the degrees of all terms...
$endgroup$
– Mindlack
Jan 5 at 22:55
$begingroup$
I fixed it now.
$endgroup$
– Jimmy Sabater
Jan 5 at 22:55
$begingroup$
I fixed it now.
$endgroup$
– Jimmy Sabater
Jan 5 at 22:55
$begingroup$
This is a very "soft" question, there is more than one answer. But the problem is that your answer keeps $B$ in it, which should be some sort of integration constant, not a known parameter of the differential equation.
$endgroup$
– orion
Jan 5 at 23:00
$begingroup$
This is a very "soft" question, there is more than one answer. But the problem is that your answer keeps $B$ in it, which should be some sort of integration constant, not a known parameter of the differential equation.
$endgroup$
– orion
Jan 5 at 23:00
$begingroup$
should I expand the binomial ?
$endgroup$
– Jimmy Sabater
Jan 5 at 23:01
$begingroup$
should I expand the binomial ?
$endgroup$
– Jimmy Sabater
Jan 5 at 23:01
$begingroup$
Your differential equation indeed has $A(x+B)^n$ as a solution. It is not, however, the only one, but I do not know whether this is relevant.
$endgroup$
– Mindlack
Jan 5 at 23:03
$begingroup$
Your differential equation indeed has $A(x+B)^n$ as a solution. It is not, however, the only one, but I do not know whether this is relevant.
$endgroup$
– Mindlack
Jan 5 at 23:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You started well, you have $y=A(x+B)^n$ and $y'=An(x+B)^{n-1}$, dividing you get
$$frac{y'}{y}=frac{n}{x+B}$$
This is already a differential equation that works for every $A$, but $B$ is still a problem (so far, you need a different differential equation for each $B$). The easiest solution I see is simply expressing it out and differentiating to kill $B$:
$$x+B=frac{y}{y'}n$$
$$1=nleft(frac{y}{y'}right)'$$
$$1=nfrac{y'^2-yy''}{y'^2}$$
$$nyy''+(1-n)y'^2=0$$
answered Jan 5 at 23:04
orionorion
13.3k11836
$endgroup$
$begingroup$
I was thinking the same but instead I did: write $y^{1/n} = A^{1/n} x + Const $ and differentiating implicitly twice gives one with no A and B
$endgroup$
– Jimmy Sabater
Jan 5 at 23:05
$begingroup$
Sure, that works too and gives you the same solution after some manipulation.
$endgroup$
– orion
Jan 5 at 23:07
add a comment |
$begingroup$
Applying Quotient rule for constant , let y= y(x)
$$ A= dfrac{y}{(x+B)^n}=dfrac{y^{'}}{n(x+B)^{n-1}} $$
Simplifying
$$dfrac{y^{'}}{ny}= dfrac{1}{x+B};quad
dfrac{ny}{y^{'}}= {x+B}; $$
Differentiate to eliminate $B$
$$ndfrac{y^{'2}-yy^{''}}{y^{'2}}=1$$
or
$$yy^{''}=y^{'2}(1-1/n) $$
which is the required second order ODE.
answered Jan 5 at 23:16
NarasimhamNarasimham
20.7k52158
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You started well, you have $y=A(x+B)^n$ and $y'=An(x+B)^{n-1}$, dividing you get
$$frac{y'}{y}=frac{n}{x+B}$$
This is already a differential equation that works for every $A$, but $B$ is still a problem (so far, you need a different differential equation for each $B$). The easiest solution I see is simply expressing it out and differentiating to kill $B$:
$$x+B=frac{y}{y'}n$$
$$1=nleft(frac{y}{y'}right)'$$
$$1=nfrac{y'^2-yy''}{y'^2}$$
$$nyy''+(1-n)y'^2=0$$
answered Jan 5 at 23:04
orionorion
13.3k11836
$endgroup$
$begingroup$
I was thinking the same but instead I did: write $y^{1/n} = A^{1/n} x + Const $ and differentiating implicitly twice gives one with no A and B
$endgroup$
– Jimmy Sabater
Jan 5 at 23:05
$begingroup$
Sure, that works too and gives you the same solution after some manipulation.
$endgroup$
– orion
Jan 5 at 23:07
add a comment |
$begingroup$
You started well, you have $y=A(x+B)^n$ and $y'=An(x+B)^{n-1}$, dividing you get
$$frac{y'}{y}=frac{n}{x+B}$$
This is already a differential equation that works for every $A$, but $B$ is still a problem (so far, you need a different differential equation for each $B$). The easiest solution I see is simply expressing it out and differentiating to kill $B$:
$$x+B=frac{y}{y'}n$$
$$1=nleft(frac{y}{y'}right)'$$
$$1=nfrac{y'^2-yy''}{y'^2}$$
$$nyy''+(1-n)y'^2=0$$
answered Jan 5 at 23:04
orionorion
13.3k11836
$endgroup$
$begingroup$
I was thinking the same but instead I did: write $y^{1/n} = A^{1/n} x + Const $ and differentiating implicitly twice gives one with no A and B
$endgroup$
– Jimmy Sabater
Jan 5 at 23:05
$begingroup$
Sure, that works too and gives you the same solution after some manipulation.
$endgroup$
– orion
Jan 5 at 23:07
add a comment |
$begingroup$
You started well, you have $y=A(x+B)^n$ and $y'=An(x+B)^{n-1}$, dividing you get
$$frac{y'}{y}=frac{n}{x+B}$$
This is already a differential equation that works for every $A$, but $B$ is still a problem (so far, you need a different differential equation for each $B$). The easiest solution I see is simply expressing it out and differentiating to kill $B$:
$$x+B=frac{y}{y'}n$$
$$1=nleft(frac{y}{y'}right)'$$
$$1=nfrac{y'^2-yy''}{y'^2}$$
$$nyy''+(1-n)y'^2=0$$
answered Jan 5 at 23:04
orionorion
13.3k11836
$endgroup$
You started well, you have $y=A(x+B)^n$ and $y'=An(x+B)^{n-1}$, dividing you get
$$frac{y'}{y}=frac{n}{x+B}$$
This is already a differential equation that works for every $A$, but $B$ is still a problem (so far, you need a different differential equation for each $B$). The easiest solution I see is simply expressing it out and differentiating to kill $B$:
$$x+B=frac{y}{y'}n$$
$$1=nleft(frac{y}{y'}right)'$$
$$1=nfrac{y'^2-yy''}{y'^2}$$
$$nyy''+(1-n)y'^2=0$$
answered Jan 5 at 23:04
orionorion
13.3k11836
answered Jan 5 at 23:04
orionorion
13.3k11836
answered Jan 5 at 23:04
orionorion
13.3k11836
answered Jan 5 at 23:04
orionorion
13.3k11836
13.3k11836
$begingroup$
I was thinking the same but instead I did: write $y^{1/n} = A^{1/n} x + Const $ and differentiating implicitly twice gives one with no A and B
$endgroup$
– Jimmy Sabater
Jan 5 at 23:05
$begingroup$
Sure, that works too and gives you the same solution after some manipulation.
$endgroup$
– orion
Jan 5 at 23:07
add a comment |
$begingroup$
I was thinking the same but instead I did: write $y^{1/n} = A^{1/n} x + Const $ and differentiating implicitly twice gives one with no A and B
$endgroup$
– Jimmy Sabater
Jan 5 at 23:05
$begingroup$
Sure, that works too and gives you the same solution after some manipulation.
$endgroup$
– orion
Jan 5 at 23:07
$begingroup$
I was thinking the same but instead I did: write $y^{1/n} = A^{1/n} x + Const $ and differentiating implicitly twice gives one with no A and B
$endgroup$
– Jimmy Sabater
Jan 5 at 23:05
$begingroup$
I was thinking the same but instead I did: write $y^{1/n} = A^{1/n} x + Const $ and differentiating implicitly twice gives one with no A and B
$endgroup$
– Jimmy Sabater
Jan 5 at 23:05
$begingroup$
Sure, that works too and gives you the same solution after some manipulation.
$endgroup$
– orion
Jan 5 at 23:07
$begingroup$
Sure, that works too and gives you the same solution after some manipulation.
$endgroup$
– orion
Jan 5 at 23:07
add a comment |
$begingroup$
Applying Quotient rule for constant , let y= y(x)
$$ A= dfrac{y}{(x+B)^n}=dfrac{y^{'}}{n(x+B)^{n-1}} $$
Simplifying
$$dfrac{y^{'}}{ny}= dfrac{1}{x+B};quad
dfrac{ny}{y^{'}}= {x+B}; $$
Differentiate to eliminate $B$
$$ndfrac{y^{'2}-yy^{''}}{y^{'2}}=1$$
or
$$yy^{''}=y^{'2}(1-1/n) $$
which is the required second order ODE.
answered Jan 5 at 23:16
NarasimhamNarasimham
20.7k52158
$endgroup$
add a comment |
$begingroup$
Applying Quotient rule for constant , let y= y(x)
$$ A= dfrac{y}{(x+B)^n}=dfrac{y^{'}}{n(x+B)^{n-1}} $$
Simplifying
$$dfrac{y^{'}}{ny}= dfrac{1}{x+B};quad
dfrac{ny}{y^{'}}= {x+B}; $$
Differentiate to eliminate $B$
$$ndfrac{y^{'2}-yy^{''}}{y^{'2}}=1$$
or
$$yy^{''}=y^{'2}(1-1/n) $$
which is the required second order ODE.
answered Jan 5 at 23:16
NarasimhamNarasimham
20.7k52158
$endgroup$
add a comment |
$begingroup$
Applying Quotient rule for constant , let y= y(x)
$$ A= dfrac{y}{(x+B)^n}=dfrac{y^{'}}{n(x+B)^{n-1}} $$
Simplifying
$$dfrac{y^{'}}{ny}= dfrac{1}{x+B};quad
dfrac{ny}{y^{'}}= {x+B}; $$
Differentiate to eliminate $B$
$$ndfrac{y^{'2}-yy^{''}}{y^{'2}}=1$$
or
$$yy^{''}=y^{'2}(1-1/n) $$
which is the required second order ODE.
answered Jan 5 at 23:16
NarasimhamNarasimham
20.7k52158
$endgroup$
Applying Quotient rule for constant , let y= y(x)
$$ A= dfrac{y}{(x+B)^n}=dfrac{y^{'}}{n(x+B)^{n-1}} $$
Simplifying
$$dfrac{y^{'}}{ny}= dfrac{1}{x+B};quad
dfrac{ny}{y^{'}}= {x+B}; $$
Differentiate to eliminate $B$
$$ndfrac{y^{'2}-yy^{''}}{y^{'2}}=1$$
or
$$yy^{''}=y^{'2}(1-1/n) $$
which is the required second order ODE.
answered Jan 5 at 23:16
NarasimhamNarasimham
20.7k52158
edited Jan 5 at 23:51
answered Jan 5 at 23:16
NarasimhamNarasimham
20.7k52158
answered Jan 5 at 23:16
NarasimhamNarasimham
20.7k52158
answered Jan 5 at 23:16
NarasimhamNarasimham
20.7k52158
20.7k52158
add a comment |
add a comment |
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$begingroup$
Your equation is wrong, check the degrees of all terms...
$endgroup$
– Mindlack
Jan 5 at 22:55
$begingroup$
I fixed it now.
$endgroup$
– Jimmy Sabater
Jan 5 at 22:55
$begingroup$
This is a very "soft" question, there is more than one answer. But the problem is that your answer keeps $B$ in it, which should be some sort of integration constant, not a known parameter of the differential equation.
$endgroup$
– orion
Jan 5 at 23:00
$begingroup$
should I expand the binomial ?
$endgroup$
– Jimmy Sabater
Jan 5 at 23:01
$begingroup$
Your differential equation indeed has $A(x+B)^n$ as a solution. It is not, however, the only one, but I do not know whether this is relevant.
$endgroup$
– Mindlack
Jan 5 at 23:03