Solve Differential Equation dy/dx = x + y for General Solution

  • Thread starter Thread starter quantum13
  • Start date Start date
quantum13
Messages
65
Reaction score
0
I read this in my calculus first year textbook, and I am just curious.

Homework Statement


For differential equation dy/dx = x + y, the general solution is y = Ce^x - x -1 (the solution of which is beyond this course). Verify this by solving dy/dx.


Homework Equations


calculus equations


The Attempt at a Solution


dy/dx = Ce^x - 1

this looks nothing like x + y however. what am I missing?
 
Physics news on Phys.org
x + y = x + Ce^x-x-1=Ce^x-1

Simply substitute your solution into the original differential equation to verify.
 
Thanks! At least it's a trivial solution, which kind of makes me feel better, but not really =]
 
I will do the same on this way:

y' = x+y

y=Ce^x - x -1

y'=Ce^x-1

Ce^x-1=y+x

y=Ce^x-1-x
 
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