You've got a few errors in your integration. Integrating -2sin6x should be 1/3 cos6x and -sin12x should be 1/12 cos12x. The final answer seems alright surprisingly.
First off, Your entire method could not possibly work if E(t) is not simply a constant, but thankfully in this case it is. You only made one simple mistake when you integrated di/[E-2i]. you're missing a -1/2 in front of the ln|E-2i|.
I approached the question by first factoring out x1/3 from the bracketted expression. with the remaining surdic expression, i rationalised the numerator in terms of cube roots so the numerator i would have
[(x2 + x + a) - (x2 - b)] = x + a + b in it.
I rearranged the x4/3 in the numerator to...
You did nothing wrong for the first problem. Just try plugging the terms into the original ODE again carefully.
For the 2nd problem, make y' the subject and think about what happens when something is implicitly differentiated. I think the answer is ey + ex + exy = 0.
Good luck.
What's so hard about this question?
It only requires some very basic trigonometric identities and properties of moduli.
Several Hints:
To derive T, 1) e-i(theta) = cos(theta) - i sin(theta)
2) How does one divide 2 complex numbers (ie. (x1 + iy1)/(x2 + iy2) )...
i know you calculate the integral to obtain the constants, but I am just at a loss in doing what you both did.
the solution to the ODE would be y = Ax + B. when inserted into the original equation,
Ax + B = 1 + (int 0->1) (x-t)y(t)dt
(A - (int 0->1)y(t)dt)x + (B - 1 + (int 0->1)...
I think you made a mistake typing it: It should be '+' in between the 2 fraction and the integral.
Just do a simple Integration by parts without induction.
Let dv=dx , u = 1/(1+x2)n
In = x/(1+x2)n + 2n(integral)[x2/(1+x2)n+1]dx
as 'x2 = 1 + x2 - 1',
You should end up with: In =...
You won't need to rely upon integrating factors in this case.
we know dy/dx = -3xy^3/(1 + 3x^2y^2)
Thus: dx/dy = -1/3xy^3 - x/y
Making a simple substitution of u = xy
dx/dy = (y*du/dy - u)/y^2 when the substitution is made
The equation should become separable.
Essentially this is using the chain rule whereby when:
G(x) = (integral of f1(x) to f2(x) ) g(t) dt
G'(x) = g(f2(x))*f2'(x) - g(f1(x))*f1'(x)
so in your case, G'(x) = sin(-x^4)*2x - sin(-x^2)
For the second question, a big hint is to equate equivalent terms.
a + bi = c + di --> a = c, b = d
Don't move things across the equals sign, but work on each side separately
what do you mean?
In this integral problem, you'll have 3 constants: one you started with (c), one you attained from the first integration (k or C1), and the last from the final integration (A)
Since you gave me pairs of values for x,T and T', the constants will result in a certain value...
Both the book and your answer is right -
though technically the answers should be 32e^-i(1/2pi).
remember, the range of principal argument is 0 < theta < pi (above the x-axis) &
-pi < theta < 0 (below the x-axis). so whenever you get a value beyond these ranges you have to convert your...
This question has quite a few twists to it, especially if you want to get an explicit equation with T as the subject. i checked my solution and you get back to the original second order equation.
dx/dT = 1/sqrt[2(k - ce^T) Where k is an arbitrary constant like you C1
to integrate this, i...
There's been one mistake running through it all... v = dT/dx NOT dx/dT
I've worked out a solution but it doesn't fit with T(1) = T(-1) = 0
But it does work if x = 0, T = 0. No problems with T'(0) = 0.
x = - sqrt(2/c) {ln|1 + sqrt(1 - e^T)| - 1/2 T}
That's the best i could do.
I got a different answer using linear differential equations.
dy/dx = 1/(x + 2y)
dx/dy = x + 2y
dx/dy - x = 2y
The answer i got was: x = -2y -2 + 6e^(y-2)
Differentiating it again returns me to the original differential
(n+3)^[1/(n+3)] is just the same as saying x^(1/x).
x^(1/x) = e^(1/x lnx)
the limit as x -> infinity, lim x^(1/x) = lim e^(1/x lnx) = e^lim (lnx / x)
lim (lnx / x) = lim (1/x) {Le Hopital's rule}
= 0
Thus, e^0 = 1
and the series is convergent