Find the general solution of a differential equation

jimkochanski
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Homework Statement



I was asked to find the general solutions of the two following differential equations:

Q1. y'''[x] + 2 y''[x] + 5 y'[x] = 0

Q2. y''[x] + 6 y'[x] + 9 y[x] = 0

Homework Equations



See above.

The Attempt at a Solution



My answers to both problems were of the form y = f[x].

My professor informed me that the general solution for the differential equation y'''[x] + 2 y''[x] + 5 y'[x] = 0 should not have been of the form y = f[x], but should have been of the form y'[x] = f[x].

Is this conventional?

Would that mean the general solution of αy(5)[x] + βy'''[x] = 0 would be of the form y'''[x] = f[x]?

Thanks for sharing your insight/feedback!

Jim
 
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The first equation does not include y(x), so it can be solved for y' as a second order de (use y'=z(x) and solve for z). After solving for z(x)=y'(x), you need to integral y' to get y. So the general solution is a function y=f(x) which contains three integration constants.

ehild
 
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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