How to solve this second-order nonlinear ordinary differential equation?

[dimension10]

I was trying to solve a calculus problem (about the equation of the curve traced by a dog chasing a postman) and I came across the following equation. I would like to know how to solve it.

\frac{w}{v}(1-\frac{y(x)y''(x)}{{y'(x)}^{2}})=\sqrt{1+{y(x)}^{2}}-\sqrt{1+{C}^{2}}

Thanks.

 

[jackmell]

I was trying to solve a calculus problem (about the equation of the curve traced by a dog chasing a postman) and I came across the following equation. I would like to know how to solve it.

\frac{w}{v}(1-\frac{y(x)y''(x)}{{y'(x)}^{2}})=\sqrt{1+{y(x)}^{2}}-\sqrt{1+{C}^{2}}

Thanks.

You have got to remove all that clutter. It just makes it way more confussing. And what's with all those extra letters. Need to absorb them into one another for now. How about start with writing it as:

k\left(1-\frac{yy''}{(y')^2}\right)=\sqrt{1+y^2}-c

See what I mean?
Now, itsn't that an equation with the independent variable missing? You know the kind where you let:

\frac{dy}{dx}=p
\frac{d^2y}{dx^2}=p\frac{dp}{dy}

Now, make all those substitutions, solve it for p in terms of y, and then integrate one more time.

 

[dimension10]

You have got to remove all that clutter. It just makes it way more confussing. And what's with all those extra letters. Need to absorb them into one another for now. How about start with writing it as:

k\left(1-\frac{yy''}{(y')^2}\right)=\sqrt{1+y^2}-c

See what I mean?
Now, itsn't that an equation with the independent variable missing? You know the kind where you let:

\frac{dy}{dx}=p
\frac{d^2y}{dx^2}=p\frac{dp}{dy}

Now, make all those substitutions, solve it for p in terms of y, and then integrate one more time.

But I tried that and got

p=\int {(\frac{dy}{dx})}^{2}(\frac{c-\sqrt{1+{y}^{2}}}{k}+1)dxwhich is equivalent to

y=\int \int {(\frac{dy}{dx})}^{2}(\frac{c-\sqrt{1+{y}^{2}}}{k}+1)dx\; dx

but I have no idea how to solve that...

 

[jackmell]

Let's solve this one:

k\left(1-\frac{yy''}{(y')^2}\right)=\sqrt{1+y^2}-c,\quad y(0)=1,\quad y'(0)=1
Letting:

\frac{dy}{dt}=p,\quad \frac{d^2y}{dy^2}=p\frac{dp}{dy}

then:

k\left(1-\frac{ypp'}{p^2}\right)=\sqrt{1+y^2}-c

at this point, separate the variables first, then let u=k+c and then integrate to obtain:

\int_{1}^{y'} k\frac{dp}{p}=\int_0^y \frac{u-\sqrt{1+y^2}}{y}dy

k\log(y')=\left(u\log(y)+\log(1+\sqrt{1+y^2})-\log(y)-\sqrt{1+y^2}\right)-k2

so that:

<br /> \begin{align*}<br /> \frac{dy}{dt}&amp;=\exp\left\{\frac{1}{k}\biggr[u\log(y)+\log(1+\sqrt{1+y^2}-\log(y)-\sqrt{1+y^2}-k2\biggr]\right\}\\<br /> &amp;=e^{f(y)}<br /> \end{align*}<br />

then we integrate again:

\int_{1}^{y} e^{-f(y)}dy=\int_0^t dt

to obtain the implicit answer:

h(y)=t

or:

y=h^{-1}(t)

Now, ain't that beautiful or what?

and we can check this numerically in Mathematica by first solving the IVP numerically, then integrate our analytic solution numerically and compare the two. Note in the Table command below, I implicitly "invert" the function h(y) by reversing the calculated values {t(y),y}:

w = 0.5; 
v = 1.7; 
k = w/v; 
c2 = 1.5; 
c = Sqrt[1 + c2^2]; 
y0 = 1; 
y1 = 1; 
u = k + c; 
k2 = Log[1 + Sqrt[2]] - Sqrt[2]

mysol = NDSolve[{k*(1 - (y[t]*Derivative[2][y][t])/Derivative[1][y][t]^2) == Sqrt[1 + y[t]^2] - c, 
    y[0] == y0, Derivative[1][y][0] == y1}, y, {t, 0, 5}]

p1 = Plot[y[t] /. mysol, {t, 0, 5}, PlotStyle -> Red]

f[y_] := Exp[(-k^(-1))*(u*Log[y] + Log[1 + Sqrt[1 + y^2]] - Log[y] - Sqrt[1 + y^2] - k2)]

myt[y_] := NIntegrate[f[s], {s, y0, y}]

mysol = Table[{myt[y], y}, {y, y0, 5, 0.1}]; 

p2 = ListPlot[mysol, Joined -> True, PlotStyle -> Blue]
Show[{p1, p2}]

There may however be a better way to solve this.

 

[dimension10]

Let's solve this one:

k\left(1-\frac{yy&#039;&#039;}{(y&#039;)^2}\right)=\sqrt{1+y^2}-c,\quad y(0)=1,\quad y&#039;(0)=1
Letting:

\frac{dy}{dt}=p,\quad \frac{d^2y}{dy^2}=p\frac{dp}{dy}

then:

k\left(1-\frac{ypp&#039;}{p^2}\right)=\sqrt{1+y^2}-c

at this point, separate the variables first, then let u=k+c and then integrate to obtain:

\int_{1}^{y&#039;} k\frac{dp}{p}=\int_0^y \frac{u-\sqrt{1+y^2}}{y}dy

k\log(y&#039;)=\left(u\log(y)+\log(1+\sqrt{1+y^2})-\log(y)-\sqrt{1+y^2}\right)-k2

so that:

<br /> \begin{align*}<br /> \frac{dy}{dt}&amp;=\exp\left\{\frac{1}{k}\biggr[u\log(y)+\log(1+\sqrt{1+y^2}-\log(y)-\sqrt{1+y^2}-k2\biggr]\right\}\\<br /> &amp;=e^{f(y)}<br /> \end{align*}<br />

then we integrate again:

\int_{1}^{y} e^{-f(y)}dy=\int_0^t dt

to obtain the implicit answer:

h(y)=t

or:

y=h^{-1}(t)

Now, ain't that beautiful or what?

and we can check this numerically in Mathematica by first solving the IVP numerically, then integrate our analytic solution numerically and compare the two. Note in the Table command below, I implicitly "invert" the function h(y) by reversing the calculated values {t(y),y}:

w = 0.5; 
v = 1.7; 
k = w/v; 
c2 = 1.5; 
c = Sqrt[1 + c2^2]; 
y0 = 1; 
y1 = 1; 
u = k + c; 
k2 = Log[1 + Sqrt[2]] - Sqrt[2]

mysol = NDSolve[{k*(1 - (y[t]*Derivative[2][y][t])/Derivative[1][y][t]^2) == Sqrt[1 + y[t]^2] - c, 
    y[0] == y0, Derivative[1][y][0] == y1}, y, {t, 0, 5}]

p1 = Plot[y[t] /. mysol, {t, 0, 5}, PlotStyle -> Red]

f[y_] := Exp[(-k^(-1))*(u*Log[y] + Log[1 + Sqrt[1 + y^2]] - Log[y] - Sqrt[1 + y^2] - k2)]

myt[y_] := NIntegrate[f[s], {s, y0, y}]

mysol = Table[{myt[y], y}, {y, y0, 5, 0.1}]; 

p2 = ListPlot[mysol, Joined -> True, PlotStyle -> Blue]
Show[{p1, p2}]

There may however be a better way to solve this.

Thanks.