Solve on (0,∞) the ODE xy'' = √{1+(y')²} with initial data y(25) = y'(25) = 0.

Answer 1

y' = u; y" = u'; xu' = √{1 + u²}; ∫du/√(1+u²) = ∫dx/x; lnIu + √1 + u²I = lnIxI + lnC; u + √1 + u² = Cx;

u(25) = 0; 1 = 25C; C = 1/25; u + √1 + u² = x/25; 1 + u² = x²/625 - 2ux/25 + u²; 2ux/25 = x²/625 - 1;

dy/dx = x/50 - 25/2x; y = x²/100 - 25lnx/2 + C; 0 = 25/4 - 25ln25/2 + C; C = 25ln25/2 - 25/4;

y = x²/100 - 25lnx/2 + 25ln25/2 - 25/4

Answer 2

The equation is nonlinear because of the square root and the dependence on y'. Our task is to find a solution for y(x) under the given conditions.

To make the equation more manageable, let’s introduce a substitution. Define: v = y'. Then, y''=dv/dx, and the equation becomes: x dv/dx = sqrt(1 + v^2)

We can now separate the variables to integrate. Rewrite the equation as: dv/sqrt(1 + v^2) = dx/x

We can now integrate both sides. The integral of dv/sqrt(1+v^2) is a standard form, which is: ∫ dv/sqrt(1 + v^2) = sinh^(-1)(v)

And the integral of dx/x is ∫ dx/x = ln|x|

So after integration, we get: sinh^(-1)(v) = ln|x| + C

Now, solve for v by taking the hyperbolic sine of both sides:

v = sinh(ln|x| + C)

Using the identity for the hyperbolic sine function, sinh(a+b)=sinh(a)cosh(b)+cosh(a)sinh(b), you can expand this, but it's simpler to treat this as: v = sinh(ln|x| + C)

^this is the expression for v=y'

We know that y'(25)=0, so 0 = sinh(ln|25| + C), which implies ln|25| + C = 0 -> C = -ln|25|

v = sinh(ln|x| - ln|25|)

v = sinh(ln(x/25))

v = (x^2 - 625) / (50x) [using sinh(ln(a))=(a^2-1)/(2a)]

Thus, y' = (x^2 - 625) / 50x

y(x) = ∫ (x^2 - 625) / (50x) dx

y(x) = (1/50) ∫ (x - 625/x) dx

y(x) = (1/50) (x^2 / 2 - 625 ln|x|) + D where D is another constant of integration

0 = (1/50) (25^2 / 2 - 625 ln|25|) + D

0 = (1/50) (625/2 - 625 ln 25) + D

D = (625/50) ln 25 - (625/100)

Thus, final solution:

y(x) = (1/50) (x^2 / 2 - 625 ln|x|) + (625/50 ln 25 - 625/100)

You can simplify this further to y = x²/100 - 25lnx/2 + 25ln25/2 - 25/4