Solving a Differential Equation with Special Integrating Factor

kuahji
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dy/dx=2+\sqrt{y-2x+3}

The only way we've been taught to solve differential equations so far is, separation of variables method, exact equation, bernoulli equation, homogeneous, and we're about to start special integrating factor.

I can't seem to get the equation into any of the first forms listed. Which leads me to believe its a special integrating factor problem. Or am I incorrect in this assumption?
 
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kuahji said:
dy/dx=2+\sqrt{y-2x+3}

The only way we've been taught to solve differential equations so far is, separation of variables method, exact equation, bernoulli equation, homogeneous, and we're about to start special integrating factor.

I can't seem to get the equation into any of the first forms listed. Which leads me to believe its a special integrating factor problem. Or am I incorrect in this assumption?

I would say try substitution v = y-2x
 
Why don't you try a change of variable v=y-2*x+3 before resorting to extreme measures? It looks dead simple to me.
 
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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