Mixing
Elementary differential equations: Please give a proof of the trick
$begingroup$
Here is a statement (or theorem) from my book:
$pmb{text{A neat trick: Turning nonlinear separable equations into linear separable equations}}$
$mathcal{text{Knowing when to substitute:}}$
You can use the trick of setting $y=xv$ when you have differential equation that is of the form:
$dfrac{dy}{dx}=f(x,y)$
when $f(x,y)=f(tx,ty)$ where $t$ is a constant.
I understand this. But the book gives no proof. I mean how can we ensure that the trick of setting $y=xv$ will always work whenever $f(x,y)=f(tx,ty)$
calculus ordinary-differential-equations substitution
asked Feb 2 at 13:37
OliverOliver
225
$endgroup$
add a comment |
$begingroup$
Here is a statement (or theorem) from my book:
$pmb{text{A neat trick: Turning nonlinear separable equations into linear separable equations}}$
$mathcal{text{Knowing when to substitute:}}$
You can use the trick of setting $y=xv$ when you have differential equation that is of the form:
$dfrac{dy}{dx}=f(x,y)$
when $f(x,y)=f(tx,ty)$ where $t$ is a constant.
I understand this. But the book gives no proof. I mean how can we ensure that the trick of setting $y=xv$ will always work whenever $f(x,y)=f(tx,ty)$
calculus ordinary-differential-equations substitution
asked Feb 2 at 13:37
OliverOliver
225
$endgroup$
$begingroup$
What have you tried ?
$endgroup$
– Yves Daoust
Feb 2 at 13:58
add a comment |
$begingroup$
Here is a statement (or theorem) from my book:
$pmb{text{A neat trick: Turning nonlinear separable equations into linear separable equations}}$
$mathcal{text{Knowing when to substitute:}}$
You can use the trick of setting $y=xv$ when you have differential equation that is of the form:
$dfrac{dy}{dx}=f(x,y)$
when $f(x,y)=f(tx,ty)$ where $t$ is a constant.
I understand this. But the book gives no proof. I mean how can we ensure that the trick of setting $y=xv$ will always work whenever $f(x,y)=f(tx,ty)$
calculus ordinary-differential-equations substitution
asked Feb 2 at 13:37
OliverOliver
225
$endgroup$
Here is a statement (or theorem) from my book:
$pmb{text{A neat trick: Turning nonlinear separable equations into linear separable equations}}$
$mathcal{text{Knowing when to substitute:}}$
You can use the trick of setting $y=xv$ when you have differential equation that is of the form:
$dfrac{dy}{dx}=f(x,y)$
when $f(x,y)=f(tx,ty)$ where $t$ is a constant.
I understand this. But the book gives no proof. I mean how can we ensure that the trick of setting $y=xv$ will always work whenever $f(x,y)=f(tx,ty)$
calculus ordinary-differential-equations substitution
calculus ordinary-differential-equations substitution
asked Feb 2 at 13:37
OliverOliver
225
asked Feb 2 at 13:37
OliverOliver
225
asked Feb 2 at 13:37
OliverOliver
225
asked Feb 2 at 13:37
OliverOliver
225
asked Feb 2 at 13:37
OliverOliver
225
225
$begingroup$
What have you tried ?
$endgroup$
– Yves Daoust
Feb 2 at 13:58
add a comment |
$begingroup$
What have you tried ?
$endgroup$
– Yves Daoust
Feb 2 at 13:58
$begingroup$
What have you tried ?
$endgroup$
– Yves Daoust
Feb 2 at 13:58
$begingroup$
What have you tried ?
$endgroup$
– Yves Daoust
Feb 2 at 13:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Just plug $y=xv$ in the equation.
$$frac{d(xv)}{dx}=xfrac{dv}{dx}+v=f(x,xv)$$
and
$$frac{dv}{f(1,v)-v}=frac{dx}x.$$
Alternatively,
$$f(tx,ty)=f(x,y)iff f(x,y)=gleft(frac yxright)=g(v)$$ and the equation is
$$xv'+v=g(v).$$
answered Feb 2 at 13:43
Yves DaoustYves Daoust
133k676231
$endgroup$
$begingroup$
If $f(x,y)=f(tx,ty)$, how can we always express $f(x,y)$ as a function of $dfrac{y}{x}$?
$endgroup$
– Oliver
Feb 2 at 15:32
$begingroup$
@Oliver: hem, $t=1/x$.
$endgroup$
– Yves Daoust
Feb 2 at 16:10
$begingroup$
Yves: THANKS FOR HELPING... However I am a bit confused here. (1) Why is $t=1/x$? (2) It is said in the question that $t$ is a constant.
$endgroup$
– Oliver
Feb 2 at 16:21
$begingroup$
@Oliver: you are right. $t$ constant is a misleading statement. Because $f(x,y)=f(tx,ty)to f(t^nx,t^ny)$ by induction and you can generalize to rational exponents. So $t$ is a weird constant...
$endgroup$
– Yves Daoust
Feb 2 at 16:26
add a comment |
Your Answer
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1 Answer
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active
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1 Answer
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active
oldest
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votes
$begingroup$
Just plug $y=xv$ in the equation.
$$frac{d(xv)}{dx}=xfrac{dv}{dx}+v=f(x,xv)$$
and
$$frac{dv}{f(1,v)-v}=frac{dx}x.$$
Alternatively,
$$f(tx,ty)=f(x,y)iff f(x,y)=gleft(frac yxright)=g(v)$$ and the equation is
$$xv'+v=g(v).$$
answered Feb 2 at 13:43
Yves DaoustYves Daoust
133k676231
$endgroup$
$begingroup$
If $f(x,y)=f(tx,ty)$, how can we always express $f(x,y)$ as a function of $dfrac{y}{x}$?
$endgroup$
– Oliver
Feb 2 at 15:32
$begingroup$
@Oliver: hem, $t=1/x$.
$endgroup$
– Yves Daoust
Feb 2 at 16:10
$begingroup$
Yves: THANKS FOR HELPING... However I am a bit confused here. (1) Why is $t=1/x$? (2) It is said in the question that $t$ is a constant.
$endgroup$
– Oliver
Feb 2 at 16:21
$begingroup$
@Oliver: you are right. $t$ constant is a misleading statement. Because $f(x,y)=f(tx,ty)to f(t^nx,t^ny)$ by induction and you can generalize to rational exponents. So $t$ is a weird constant...
$endgroup$
– Yves Daoust
Feb 2 at 16:26
add a comment |
$begingroup$
Just plug $y=xv$ in the equation.
$$frac{d(xv)}{dx}=xfrac{dv}{dx}+v=f(x,xv)$$
and
$$frac{dv}{f(1,v)-v}=frac{dx}x.$$
Alternatively,
$$f(tx,ty)=f(x,y)iff f(x,y)=gleft(frac yxright)=g(v)$$ and the equation is
$$xv'+v=g(v).$$
answered Feb 2 at 13:43
Yves DaoustYves Daoust
133k676231
$endgroup$
$begingroup$
If $f(x,y)=f(tx,ty)$, how can we always express $f(x,y)$ as a function of $dfrac{y}{x}$?
$endgroup$
– Oliver
Feb 2 at 15:32
$begingroup$
@Oliver: hem, $t=1/x$.
$endgroup$
– Yves Daoust
Feb 2 at 16:10
$begingroup$
Yves: THANKS FOR HELPING... However I am a bit confused here. (1) Why is $t=1/x$? (2) It is said in the question that $t$ is a constant.
$endgroup$
– Oliver
Feb 2 at 16:21
$begingroup$
@Oliver: you are right. $t$ constant is a misleading statement. Because $f(x,y)=f(tx,ty)to f(t^nx,t^ny)$ by induction and you can generalize to rational exponents. So $t$ is a weird constant...
$endgroup$
– Yves Daoust
Feb 2 at 16:26
add a comment |
$begingroup$
Just plug $y=xv$ in the equation.
$$frac{d(xv)}{dx}=xfrac{dv}{dx}+v=f(x,xv)$$
and
$$frac{dv}{f(1,v)-v}=frac{dx}x.$$
Alternatively,
$$f(tx,ty)=f(x,y)iff f(x,y)=gleft(frac yxright)=g(v)$$ and the equation is
$$xv'+v=g(v).$$
answered Feb 2 at 13:43
Yves DaoustYves Daoust
133k676231
$endgroup$
Just plug $y=xv$ in the equation.
$$frac{d(xv)}{dx}=xfrac{dv}{dx}+v=f(x,xv)$$
and
$$frac{dv}{f(1,v)-v}=frac{dx}x.$$
Alternatively,
$$f(tx,ty)=f(x,y)iff f(x,y)=gleft(frac yxright)=g(v)$$ and the equation is
$$xv'+v=g(v).$$
answered Feb 2 at 13:43
Yves DaoustYves Daoust
133k676231
answered Feb 2 at 13:43
Yves DaoustYves Daoust
133k676231
answered Feb 2 at 13:43
Yves DaoustYves Daoust
133k676231
answered Feb 2 at 13:43
Yves DaoustYves Daoust
133k676231
133k676231
$begingroup$
If $f(x,y)=f(tx,ty)$, how can we always express $f(x,y)$ as a function of $dfrac{y}{x}$?
$endgroup$
– Oliver
Feb 2 at 15:32
$begingroup$
@Oliver: hem, $t=1/x$.
$endgroup$
– Yves Daoust
Feb 2 at 16:10
$begingroup$
Yves: THANKS FOR HELPING... However I am a bit confused here. (1) Why is $t=1/x$? (2) It is said in the question that $t$ is a constant.
$endgroup$
– Oliver
Feb 2 at 16:21
$begingroup$
@Oliver: you are right. $t$ constant is a misleading statement. Because $f(x,y)=f(tx,ty)to f(t^nx,t^ny)$ by induction and you can generalize to rational exponents. So $t$ is a weird constant...
$endgroup$
– Yves Daoust
Feb 2 at 16:26
add a comment |
$begingroup$
If $f(x,y)=f(tx,ty)$, how can we always express $f(x,y)$ as a function of $dfrac{y}{x}$?
$endgroup$
– Oliver
Feb 2 at 15:32
$begingroup$
@Oliver: hem, $t=1/x$.
$endgroup$
– Yves Daoust
Feb 2 at 16:10
$begingroup$
Yves: THANKS FOR HELPING... However I am a bit confused here. (1) Why is $t=1/x$? (2) It is said in the question that $t$ is a constant.
$endgroup$
– Oliver
Feb 2 at 16:21
$begingroup$
@Oliver: you are right. $t$ constant is a misleading statement. Because $f(x,y)=f(tx,ty)to f(t^nx,t^ny)$ by induction and you can generalize to rational exponents. So $t$ is a weird constant...
$endgroup$
– Yves Daoust
Feb 2 at 16:26
$begingroup$
If $f(x,y)=f(tx,ty)$, how can we always express $f(x,y)$ as a function of $dfrac{y}{x}$?
$endgroup$
– Oliver
Feb 2 at 15:32
$begingroup$
If $f(x,y)=f(tx,ty)$, how can we always express $f(x,y)$ as a function of $dfrac{y}{x}$?
$endgroup$
– Oliver
Feb 2 at 15:32
$begingroup$
@Oliver: hem, $t=1/x$.
$endgroup$
– Yves Daoust
Feb 2 at 16:10
$begingroup$
@Oliver: hem, $t=1/x$.
$endgroup$
– Yves Daoust
Feb 2 at 16:10
$begingroup$
Yves: THANKS FOR HELPING... However I am a bit confused here. (1) Why is $t=1/x$? (2) It is said in the question that $t$ is a constant.
$endgroup$
– Oliver
Feb 2 at 16:21
$begingroup$
Yves: THANKS FOR HELPING... However I am a bit confused here. (1) Why is $t=1/x$? (2) It is said in the question that $t$ is a constant.
$endgroup$
– Oliver
Feb 2 at 16:21
$begingroup$
@Oliver: you are right. $t$ constant is a misleading statement. Because $f(x,y)=f(tx,ty)to f(t^nx,t^ny)$ by induction and you can generalize to rational exponents. So $t$ is a weird constant...
$endgroup$
– Yves Daoust
Feb 2 at 16:26
$begingroup$
@Oliver: you are right. $t$ constant is a misleading statement. Because $f(x,y)=f(tx,ty)to f(t^nx,t^ny)$ by induction and you can generalize to rational exponents. So $t$ is a weird constant...
$endgroup$
– Yves Daoust
Feb 2 at 16:26
add a comment |
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$begingroup$
What have you tried ?
$endgroup$
– Yves Daoust
Feb 2 at 13:58