ODE5.4

From Example Problems
Jump to: navigation, search

{\frac  {1}{y}}={\frac  {y''}{1+y'^{2}}}

Multiply both sides by 2y'\,:

{\frac  {2y'}{y}}={\frac  {2y'y''}{1+y'^{2}}}

Observe that the numerator of the term on the right side is the derivative (with respect to x) of the denominator of that term (indeed, that is why the factor 2y' was chosen -- a sort of integrating factor). Integrate both sides with respect to x; the result is

2\ln |y|=\ln(1+y'^{2})+a\,

(Alternatively, since the differential equation does not explicitly involve x, the substitution {\frac  {dy}{dx}}=p,{\frac  {d^{2}y}{dx^{2}}}=p{\frac  {dp}{dy}} converts the equation to a first-order equation in p and y which is separable. Integrating and back-substituting dy/dx\, for p\, yields the same result as given above, as of course it must.)

Exponentiating both sides, we gety^{2}=c^{2}(1+y'^{2})\,

where c^{2}=e^{a}\, (we are free to write c^{2} since e^{a}>0 for all real a).

Solving the above for y',

y'=\pm {\frac  {{\sqrt  {y^{2}-c^{2}}}}{c}}

Divide both sides by {\sqrt  {y^{2}-c^{2}}} and integrate:

\int {\frac  {dy}{{\sqrt  {y^{2}-c^{2}}}}}=\int \pm {\frac  {dx}{c}} (where we have used y'dx=dy in the integral on the left)

Use the trig substitution y=c\,\sec(\theta ), dy=c\,\sec(\theta )\tan(\theta )\,d\theta to get

\int \sec(\theta )\,d\theta =\pm \int {\frac  {dx}{c}}\qquad Integrating,

\ln {|\sec(\theta )+\tan(\theta )|}=\pm x/c+d\, or after back-substitution,

\ln {|y/c+{\sqrt  {y^{2}/c^{2}-1}}|}=\pm x/c+d\qquad Exponentiating both sides and multiplying both sides by c,

y+{\sqrt  {y^{2}-1}}=ce^{{\pm (x/c)+d}}\qquad Solving this for y,

y={\frac  {c\left(e^{{2(\pm (x/c)+d)}}+1\right)}{2e^{{\pm (x/c)+d}}}}=c\,{\frac  {e^{{\pm (x/c)+d}}+e^{{\mp (x/c)-d}}}{2}}

Either case may be written in the form y=c\cosh(x/c\,+\,b)\, where b and c are arbitrary constants, as may be verified by substituting this solution into the original differential equation.

Ordinary Differential Equations

Main Page