Solve. x^2y'' + xy' + \left(x^2 - \frac{25}{4}\right)y = 0
logistic_guy Senior Member Joined Apr 17, 2024 Messages 1,132 Jun 26, 2025 #1 Solve. x2y′′+xy′+(x2−254)y=0\displaystyle x^2y'' + xy' + \left(x^2 - \frac{25}{4}\right)y = 0x2y′′+xy′+(x2−425)y=0
Solve. x2y′′+xy′+(x2−254)y=0\displaystyle x^2y'' + xy' + \left(x^2 - \frac{25}{4}\right)y = 0x2y′′+xy′+(x2−425)y=0
logistic_guy Senior Member Joined Apr 17, 2024 Messages 1,132 Jun 28, 2025 #2 logistic_guy said: Solve. x2y′′+xy′+(x2−254)y=0\displaystyle x^2y'' + xy' + \left(x^2 - \frac{25}{4}\right)y = 0x2y′′+xy′+(x2−425)y=0 Click to expand... We all now can solve this differential equation quickly. With full of confidence we can say that the general solution is: y(x)=c1J5/2(x)+c2J−5/2(x)\displaystyle y(x) = c_1J_{5/2}(x) + c_2J_{-5/2}(x)y(x)=c1J5/2(x)+c2J−5/2(x) Writing this solution in terms of trigonometric functions is a little bit challenging. But we love Challenges We start with our secret weapon which is the recurrence relation. 2nJn(x)=xJn+1(x)+xJn−1(x)\displaystyle 2nJ_n(x) = xJ_{n + 1}(x) + xJ_{n - 1}(x)2nJn(x)=xJn+1(x)+xJn−1(x) This time let us try n=32\displaystyle n = \frac{3}{2}n=23 and see what we get. 232J3/2(x)=xJ3/2+1(x)+xJ3/2−1(x)\displaystyle 2\frac{3}{2}J_{3/2}(x) = xJ_{3/2 + 1}(x) + xJ_{3/2 - 1}(x)223J3/2(x)=xJ3/2+1(x)+xJ3/2−1(x) 3J3/2(x)=xJ5/2(x)+xJ1/2(x)\displaystyle 3J_{3/2}(x) = xJ_{5/2}(x) + xJ_{1/2}(x)3J3/2(x)=xJ5/2(x)+xJ1/2(x) This gives: J5/2(x)=3xJ3/2(x)−J1/2(x)\displaystyle J_{5/2}(x) = \frac{3}{x}J_{3/2}(x) - J_{1/2}(x)J5/2(x)=x3J3/2(x)−J1/2(x) From previous exercises, we know that: J3/2(x)=1x2πxsinx−2πxcosx\displaystyle J_{3/2}(x) = \frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos xJ3/2(x)=x1πx2sinx−πx2cosx And J1/2(x)=2πxsinx\displaystyle J_{1/2}(x) = \sqrt{\frac{2}{\pi x}}\sin xJ1/2(x)=πx2sinx Then J5/2(x)=3x(1x2πxsinx−2πxcosx)−2πxsinx\displaystyle J_{5/2}(x) = \frac{3}{x}\left(\frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\sin xJ5/2(x)=x3(x1πx2sinx−πx2cosx)−πx2sinx Again let us try n=−32\displaystyle n = -\frac{3}{2}n=−23 and see what that leads us. −232J−3/2(x)=xJ−3/2+1(x)+xJ−3/2−1(x)\displaystyle -2\frac{3}{2}J_{-3/2}(x) = xJ_{-3/2 + 1}(x) + xJ_{-3/2 - 1}(x)−223J−3/2(x)=xJ−3/2+1(x)+xJ−3/2−1(x) −3J−3/2(x)=xJ−1/2(x)+xJ−5/2(x)\displaystyle -3J_{-3/2}(x) = xJ_{-1/2}(x) + xJ_{-5/2}(x)−3J−3/2(x)=xJ−1/2(x)+xJ−5/2(x) This gives: J−5/2(x)=−3xJ−3/2(x)−J−1/2(x)\displaystyle J_{-5/2}(x) = -\frac{3}{x}J_{-3/2}(x) - J_{-1/2}(x)J−5/2(x)=−x3J−3/2(x)−J−1/2(x) From previous exercises, we know that: J−3/2(x)=−2πxsinx−1x2πxcosx\displaystyle J_{-3/2}(x) = -\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos xJ−3/2(x)=−πx2sinx−x1πx2cosx And J−1/2(x)=2πxcosx\displaystyle J_{-1/2}(x) = \sqrt{\frac{2}{\pi x}}\cos xJ−1/2(x)=πx2cosx Then J−5/2(x)=−3x(−2πxsinx−1x2πxcosx)−2πxcosx\displaystyle J_{-5/2}(x) = -\frac{3}{x}\left(-\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\cos xJ−5/2(x)=−x3(−πx2sinx−x1πx2cosx)−πx2cosx Now we can write the general solution in terms of trigonometric functions. y(x)=c1[3x(1x2πxsinx−2πxcosx)−2πxsinx]+c2[−3x(−2πxsinx−1x2πxcosx)−2πxcosx]\displaystyle y(x) = c_1\left[\frac{3}{x}\left(\frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\sin x\right] + c_2\left[-\frac{3}{x}\left(-\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\cos x\right]y(x)=c1[x3(x1πx2sinx−πx2cosx)−πx2sinx]+c2[−x3(−πx2sinx−x1πx2cosx)−πx2cosx] We can simplify it further if we want in terms of the arbitrary constants C1,C2,C3,C4,C5,and C6\displaystyle C_1, C_2, C_3, C_4, C_5, \text{and} \ C_6C1,C2,C3,C4,C5,and C6. I think that you got the idea!
logistic_guy said: Solve. x2y′′+xy′+(x2−254)y=0\displaystyle x^2y'' + xy' + \left(x^2 - \frac{25}{4}\right)y = 0x2y′′+xy′+(x2−425)y=0 Click to expand... We all now can solve this differential equation quickly. With full of confidence we can say that the general solution is: y(x)=c1J5/2(x)+c2J−5/2(x)\displaystyle y(x) = c_1J_{5/2}(x) + c_2J_{-5/2}(x)y(x)=c1J5/2(x)+c2J−5/2(x) Writing this solution in terms of trigonometric functions is a little bit challenging. But we love Challenges We start with our secret weapon which is the recurrence relation. 2nJn(x)=xJn+1(x)+xJn−1(x)\displaystyle 2nJ_n(x) = xJ_{n + 1}(x) + xJ_{n - 1}(x)2nJn(x)=xJn+1(x)+xJn−1(x) This time let us try n=32\displaystyle n = \frac{3}{2}n=23 and see what we get. 232J3/2(x)=xJ3/2+1(x)+xJ3/2−1(x)\displaystyle 2\frac{3}{2}J_{3/2}(x) = xJ_{3/2 + 1}(x) + xJ_{3/2 - 1}(x)223J3/2(x)=xJ3/2+1(x)+xJ3/2−1(x) 3J3/2(x)=xJ5/2(x)+xJ1/2(x)\displaystyle 3J_{3/2}(x) = xJ_{5/2}(x) + xJ_{1/2}(x)3J3/2(x)=xJ5/2(x)+xJ1/2(x) This gives: J5/2(x)=3xJ3/2(x)−J1/2(x)\displaystyle J_{5/2}(x) = \frac{3}{x}J_{3/2}(x) - J_{1/2}(x)J5/2(x)=x3J3/2(x)−J1/2(x) From previous exercises, we know that: J3/2(x)=1x2πxsinx−2πxcosx\displaystyle J_{3/2}(x) = \frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos xJ3/2(x)=x1πx2sinx−πx2cosx And J1/2(x)=2πxsinx\displaystyle J_{1/2}(x) = \sqrt{\frac{2}{\pi x}}\sin xJ1/2(x)=πx2sinx Then J5/2(x)=3x(1x2πxsinx−2πxcosx)−2πxsinx\displaystyle J_{5/2}(x) = \frac{3}{x}\left(\frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\sin xJ5/2(x)=x3(x1πx2sinx−πx2cosx)−πx2sinx Again let us try n=−32\displaystyle n = -\frac{3}{2}n=−23 and see what that leads us. −232J−3/2(x)=xJ−3/2+1(x)+xJ−3/2−1(x)\displaystyle -2\frac{3}{2}J_{-3/2}(x) = xJ_{-3/2 + 1}(x) + xJ_{-3/2 - 1}(x)−223J−3/2(x)=xJ−3/2+1(x)+xJ−3/2−1(x) −3J−3/2(x)=xJ−1/2(x)+xJ−5/2(x)\displaystyle -3J_{-3/2}(x) = xJ_{-1/2}(x) + xJ_{-5/2}(x)−3J−3/2(x)=xJ−1/2(x)+xJ−5/2(x) This gives: J−5/2(x)=−3xJ−3/2(x)−J−1/2(x)\displaystyle J_{-5/2}(x) = -\frac{3}{x}J_{-3/2}(x) - J_{-1/2}(x)J−5/2(x)=−x3J−3/2(x)−J−1/2(x) From previous exercises, we know that: J−3/2(x)=−2πxsinx−1x2πxcosx\displaystyle J_{-3/2}(x) = -\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos xJ−3/2(x)=−πx2sinx−x1πx2cosx And J−1/2(x)=2πxcosx\displaystyle J_{-1/2}(x) = \sqrt{\frac{2}{\pi x}}\cos xJ−1/2(x)=πx2cosx Then J−5/2(x)=−3x(−2πxsinx−1x2πxcosx)−2πxcosx\displaystyle J_{-5/2}(x) = -\frac{3}{x}\left(-\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\cos xJ−5/2(x)=−x3(−πx2sinx−x1πx2cosx)−πx2cosx Now we can write the general solution in terms of trigonometric functions. y(x)=c1[3x(1x2πxsinx−2πxcosx)−2πxsinx]+c2[−3x(−2πxsinx−1x2πxcosx)−2πxcosx]\displaystyle y(x) = c_1\left[\frac{3}{x}\left(\frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\sin x\right] + c_2\left[-\frac{3}{x}\left(-\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\cos x\right]y(x)=c1[x3(x1πx2sinx−πx2cosx)−πx2sinx]+c2[−x3(−πx2sinx−x1πx2cosx)−πx2cosx] We can simplify it further if we want in terms of the arbitrary constants C1,C2,C3,C4,C5,and C6\displaystyle C_1, C_2, C_3, C_4, C_5, \text{and} \ C_6C1,C2,C3,C4,C5,and C6. I think that you got the idea!