Differential Equation help 1.?

Question

Find the general solution to the DE.

y'=(2x+3y)/x Thanks for any help!

Answer 1

This is a linear differential equation. First we subtract 3y/x to get it in standard form:

y'-3/x y=2

Now, we multiply by an integrating factor of e^(∫-3/x dx) = e^(-3 ln x) = x^(-3):

x^(-3) y' - 3x^(-4) y = 2x^(-3)

Now we integrate both sides, using the inverse product rule on the left:

x^(-3)y = -x^(-2) + C

Finally, we isolate y by multiplying by x^3:

y = -x+Cx^3

And we are done.

Answer 2

xy' - 3y = 2x

You could divide this problem into two solutions.
1) one that gives you xy' - 3y = 0
2) the other that gives xy' - 3y = 2x

1) Because the derivative is being multiplied by a higher power of x, we can assume a polynomial form
y = Cx^n
y' = n*Cx^)n-1)
So xy' + y = C*(n - 3)*x^n = 0
This happens if n = 3.
So y = Cx^3 is one solution

2) To get the RHS = 2x
assume y = Ax + B
y' = A
So xy' - 3x = Ax - 3Ax - 3B = 2x
-2Ax - 3B = 2x
clearly A = -1, B = 0
so this solution is y = -x

FULL solution y = Cx^3 - x

Answer 3

hey kayla, are their any conditions to this differential equation??
other than that, you cant solve it.