Re: Proving a formula is a solution

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: : : : I am trying to verify that the given formula is a solution to the initial value problem. : : : : Y'=y^3 : : : : y(0) = 5 : : : : Formula: y(t) = 1 / sqrt(1/25-2t) : : : : I know that 1/sqrt(1/25-2t) at y(0) = 5 : : : : When I try to differentiate this, though, I run into problems. : : : : -(sqrt 1/25-2t)^-2* [(1/25-2t)^-1/2]/2 * 2 : : : : Since the twos cancel, I end up with : : : : -(sqrt 1/25-2t)^-2* [(1/25-2t)^-1/2] : : : : That can't be right since I am sure this formula is a solution - just looking at the problems that follow this. What have I done wrong? : : : That's very confusing & I can't really figure out exactly what you did. Obviously you are using the quotient rule, which will work, but since the numerator is just a constant why not make your life easier & rewrite the function as: : : : f(t) = (1/25 - 2t)^(-1/2) : : : Now: : : : f'(t) = (-2)*(-1/2) * (1/25-2t)^(-3/2) : : : ..... = (1/25-2t)^(-3/2) : : : and that's all there is to it. : : : If you insist on the quotient rule: : : : f'(t) = [-(-2)(1/2)(1/25-2t)^(-1/2)] / [(sqrt(1/25-2t))^2] : : : which eventually simplifies to the same thing. : : So then this is not a solution to the original derivative y'=y^3 since the two derivatives do not equal one another? : One more step: : f(t) = 1 / sqrt(1/25-2t) (given) : .... = (1/25-2t)^(-1/2) (per above) : raise it to 3rd power: : (f(t))^3 = (1/25-2t)^(-3/2) (same as f'(t))

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