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y''-10y'+25y=0 solve using reduction of order with one solution being y=e^5x
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d^2 - 10d+25 = 0, d=5, -5, the other solution would be e^-5x
How did you get this? I need to know how to solve the problem
i have little understanding on this, but somehow second order differentiantion is done this way d stands for dy/dx and d^2 for second order. you find the complimentary function which is equal to the e^mx where m is root of above quadratic
and the general solution would be c1 e^5x + c2 e^-5x but never ask me how??
So i"m using reduction of order method and I'm stuck at v''e^5x=0 I don't know where to go after this step
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