Solve DE: Find Solution for y Through (-1,-1)

  • Thread starter Thread starter ssb
  • Start date Start date
ssb
Messages
119
Reaction score
0

Homework Statement



\frac{dy}{dt}=y\cos(t)

Find the solution of the DE that passes through the point (-1, -1).

Homework Equations


The Attempt at a Solution



\frac{dy}{dt}=y\cos(t)

\frac{1}{y}dy=cos(t)dt

integrate both sides:

ln(y) = sin(t) + C

Normally I would plug in -1 for y and t and solve for C but I can't take the LN of -1. When I try to isolate y first then plug in, I get the same problem with t. How can I solve this? I am stuck! Is it that the ln(y) is actually ln(|y|) ?
 
Physics news on Phys.org
You figured it out for yourself in the end:smile::

\int\frac{dy}{y}=\ln|y|
 
Last edited:
You could do the absolute value, since ln(-y)'=ln(y)=1/y*y'. Or you could just exponentiate your solution to get y=exp(sin(t)+C)=D*exp(sin(t)). Now you can put D negative.
 
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