Solve a first-order nonlinear ordinary differential equation (boundary value problem)I have been trying to solve the following boundary value problem:(x2−36)(y′)2−2xyy′+y2−36=0Given the conditions y(0)=6 and y(10)=0, where y(x) is continuously twice differentiable on [0,10]. The requirement is not really to solve the problem itself, but to find∫010y(x)dxI have almost found the answer, but got stuck. My idea was to solve for y and not for y':y=xy′±6(y′)2+1And then to directly integrate this expression. Finally, I ended up with:∫010y(x)dx=±3∫010((y′)2+1)dxBut I do not understand how to solve the integral on the right, I have tried many substitutions, but they led nowhere. This leads me to believe that my overall idea to solve for y was a mistake, but I am not sure. I have tried solving for y' and for x, but this did not help either. I would appreciate any suggestions on how to solve the problem.

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Solve a first-order nonlinear ordinary differential equation (boundary value problem)I have been trying to solve the following boundary value problem:(x2−36)(y′)2−2xyy′+y2−36=0Given the conditions y(0)=6 and y(10)=0, where y(x) is continuously twice differentiable on [0,10]. The requirement is not really to solve the problem itself, but to find∫010y(x)dxI have almost found the answer, but got stuck. My idea was to solve for y and not for y&#039;:y=xy′±6(y′)2+1And then to directly integrate this expression. Finally, I ended up with:∫010y(x)dx=±3∫010((y′)2+1)dxBut I do not understand how to solve the integral on the right, I have tried many substitutions, but they led nowhere. This leads me to believe that my overall idea to solve for y was a mistake, but I am not sure. I have tried solving for y&#039; and for x, but this did not help either. I would appreciate any suggestions on how to solve the problem.

Jocelynn Meyer · Accepted Answer

Step 1Add the derivative of the equation to the set of knowledge[2y′(x2−36)−2xy]y″=0This means that solutions can have segments where they are linear,y=Cx+D,where the constants may be connected via the original equation, and other segments whereStep 2y′y=xx2−36⇒y2=C(x2−36),where again C might be bound by the original equation.Insert these formulas and solve for the constants. Then try to find the correct combination of such pieces to satisfy the boundary condition. It might not be possible to get an everywhere continuous second derivative.That the derivative factors so well is related to the fact that the reduction of the first equation via quadratic equation solution formula results in Clairaut equations.

Solve a first-order nonlinear ordinary differential equation (boundary value problem) I have bee

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