Differential Equation- Separation of Variables

brutalmadness
Messages
16
Reaction score
0

Homework Statement


y'=xsec^2(x^2)

2. The attempt at a solution
dy/dx=xsec^2(x^2)
dy=xsec^2(x^2)dx
\intdy=\intxsec^2(x^2)dx
lny= (here i'll do a u substitution)
----
u=x^2 du=1/3x^3dx


... and here's my problem. It seems like that creates a very difficult u-sub to try and manage. Any suggestions/help?
 
Physics news on Phys.org
brutalmadness said:
lny= (here i'll do a u substitution)
----
u=x^2 du=1/3x^3dx
This is incorrect. Your method is correct, though.
 
I think you should write sec function in terms of cos. Function will be dy = x / (cos^2(x^2)). Then if you say x^2 = u and du = 2x.dx, you can solve the problem easily.
 
u=x^2 du=2xdx
2\intdu/cos^2(u)
 
You're off by a factor of 1/4. It should be \frac{1}{2} \int sec^2u \ du.
 
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