find the general solution of (dz/dx) - (2z/x) = 2/3 * x^4?

Answer 1

The general solution is:
z=2(x^5)/9 + C(x^2)

Follow exactly the same procedure as I explained in my PDF file (which I posted for your other question).

Take the following changes into account:
P(x)=(-2/x)
Q(x)=2*(x^4)/3

I.F. will turn out to be 1/x^2

Answer 2

(-2z/x) = (2/3)x^4
(-2z/x)(-x/2) = (-x/2)(2/3)x^4
z = (-1/3)x^5
For the derivative
The exponent ^5 becomes a factor 5.
The -1/3 remains a factor.
Subtract 1 from the exponent ^5.
(-5/3)x^4 = dz/dx

Answer 3

Solving this problem online is rather cumbersome. i give you the basic method. This is a linear first order diff eqn in z & x. These kind of eqns have the general form:
[dz/dx]+ z*p(x)=q(x), p& q being 2 fns of x. Here we multiply throughout by e^[integral p(x)dx] converting it to the form:
z*e^[integral p(x)dx]= integral[q(x)*e^[integral p(x)dx]dx
Now u should b able to complete the problem. BEST OF LUCK

Answer 4

$50

Answer 5

Rewrite:
z'-2x*z=(2/3)x^4

First order linear differential, use the integrating factor.

I(x)=e^int(-2x*dx)=e^(-x^2)

Multiply through by I(x)
e^(-x^2)*z'-2x*e^(-x^2)*z=(2/3)x^4*e^(-x^2)

Compact left hand side into a chain rule derivative:
d/dx(e^(-x^2)*z)=(2/3)x^4*e^(-x^2)

Integrate both sides wrt x:

e^(-x^2)*z=(2/3)*int(x^4*e^(-x^2)*dx)

Solve for z:

z=((2/3)/(e^(-x^2)))*int(x^4*e^(-x^2)*dx)

To solve the integral, you must use a couple of steps of U*dV substitution. I'm going to cheat b/c I am lazy....(hint, let u=polynomial):

z=((2/3)/(e^(-x^2)))*[(-1/2)*(x^3)/(e^(x^2) - (3/4)*(x)/(e^x^2) + (3/8)*sqrt(pi)*erf(x)]

or

z=[(-1/3)*(x^3) - (1/2)*(x) + (1/4)*sqrt(pi)*e^(x^2)*erf(x)]

Answer 6

666

Answer 7

first tell what is dz dx 2z x and^

Answer 8

2!!

Answer 9

-3z=2x^5