Differential Equations Problem

Dao Tuat
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Could someone please help me with this problem:

Consider the differential equation d^2y/dx^2 + dy/dx - 6y = 0 with the initial conditions y(0)= 6 and y'(0)=2. Determine whether the following functions are solutions:

a. y=4e^(2x) + 2e^(-3x)

b. y=2e^(2x) + 4e^(-3x)

If someone could please at least help me get started on this I would appreciate it very much

Thanks,
Dao Tuat
 
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You need to substitute the two solutions given into the differential equation, and see if they come to zero. I don't think they're asking you to solve it directly.
 
Be sure to also check that the solutions satisfy the given initial conditions.
 
So then neither a or b are solutions, right?
 
Only a. is a solution because b. doesn't satisfy y'(0) = 2 (y'(0) = -8). They both fit the equation, though, as can seen below.

(16e^(2x) + 18e^(-3x)) + (8e^(2x) - 6e^(2x)) - 6*(4e^(2x) + 2e^(-3x)
(8e^(2x) + 36e^(-3x)) + (4e^(2x) - 12e^(-3x)) - 6*(2e^(2x) + 4e^(-3x))
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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