Verifying a solution to a DE (please check my work)

[darryw]

Homework Statement

Is y_1 = 1 and y_2 = root t solutions of the eqn: yy'' + (y')^2 = 0 ?

first solution works (i already verified)

2nd solution i get this:

(1/2)t^(-1) + (1/4)t^(1/4) which does not equal zero.

is this correct so far? the thing that confuses me is the question tells me to "verfiy that they are solutions" but one of them isnt. Either i mutiplied exponents incorrectly, or y_2 is not a solution.
thanks for any help.

Homework Equations

The Attempt at a Solution

 

[Integral]

Show us the 1st and 2nd derivatives of \sqrt {t}

 

[darryw]

y(t) = t^(1/2)

y'(t) = (1/2)t^(-1/2)

y''(t) = (1/4)t^(-3/2)

plug into equation..

t^(1/2)((1/4)t^(-3/2) + ( (1/2)t^(-1/2))^2

(1/2)t^(-1) + (1/4)t^(1/4)

thanks

 

[darryw]

hang on sec.. i see something stupid

 

[darryw]

it was property of exponents that was problem.. wow.
so y_2 = root t is also a solution

(1/4t) - (1/4t) = 0

 

[Integral]

y(t) = t^(1/2)

y'(t) = (1/2)t^(-1/2)

y''(t) = (1/4)t^(-3/2)

You also have an issure withe the 2nd der. You lost a minus sign, it should read:
y"(t) = - \frac 1 4 t^{- \frac 3 2}
Now it should all work out.

plug into equation..

t^(1/2)((1/4)t^(-3/2) + ( (1/2)t^(-1/2))^2

(1/2)t^(-1) + (1/4)t^(1/4)

thanks

 

[Mark44]

it was property of exponents that was problem.. wow.
so y_2 = root t is also a solution

(1/4t) - (1/4t) = 0

One other thing. (1/4t) is usually interpreted to mean (1/4) * t. If you want t in the denominator, write this as 1/(4t).

 

[darryw]

next part asks about linear independence, but i already know this is just making sure wronskian doesn't equal zero. thanks for all the help