Solution of Initial Value Problem

DrunkApple
Messages
110
Reaction score
0

Homework Statement


Determine the solution of the IVP y' + 4ty = 4t, y(0) = 6


Homework Equations





The Attempt at a Solution


p(t) = 4t
g(t) = 4t

μ(t) = e^{\int4tdt}
= e^{\int p(t)}
= e^{\int4tdt}
= e^{2t^{2}}

is this all I need? because i did
\frac{d}{dt}(y * μ(t)) = p(t) * g(t)
and the professor made a big red X mark on it, so I am confused.
 
Physics news on Phys.org
Hi DrunkApple! :smile:
DrunkApple said:
p(t) = 4t
g(t) = 4t

\frac{d}{dt}(y * μ(t)) = p(t) * g(t)

nooo :redface:

you're multiplying the whole equation by e2t2,

so the RHS should be e2t2 * g(t), shouldn't it? :wink:
 
wow...
can't believe i made that mistake...
thank you super hero tim-tim
 
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...
Back
Top