Find integrating factor and solve the equation 3

naspek
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y' + y = e^x ; y(0) = 1

1st, i calculate the integrating factor...
u(x) = e^x

times the integrating factor with DE...

y'e^x + ye^x = e^2x

dy/dx e^x + ye^x = e^2x

d/dx ye^x = e^2x

ye^x = ∫ e^2x dx
...= 1/2 e^2x + C

y = 1/2 e^x + C

the problem here, i didn't get the answer given which is..
y = 1/2 (e^x + e^-x)
 
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You made an algebra mistake in the last step when solving for y.
 
ok.. here is my mistake...
ye^x = ∫ e^2x dx
...= 1/2 e^2x + C

so.. when am i going to solve C value?
 
naspek said:
ok.. here is my mistake...
ye^x = ∫ e^2x dx
...= 1/2 e^2x + C

so.. when am i going to solve C value?

Up to there is correct. Your error was in the very last step you wrote when you solved for y.

You can solve for C any time you want. Most of the time, it's done as the final step.
 
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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