Differential Equation, DE

Differential Equation: $y' = x^3 - 2xy$, where $y(1)=1$ and $y' = 2(2x-y)$ that passes through (0,1)

Can anyone solve this D. E.?

y' = x^3 - 2xy, where y(1)=1

and

y' = 2(2x-y) that passes through (0,1)

Joshua Melegrito Peralta's picture

Help please: $(1+e^x y+x e^x y) dx + (x e^x + 2) dy=0$

Help please.
(1+e^(x) y+x e^(x) y) dx + (x e^(x) + 2) dy=0

Arbitrary constant

y=x^2+C1e^2x+C2e^3x

Thank you again

Families of Curves: family of circles with center on the line y= -x and passing through the origin

Find the diff equation of family of circles with center on the line y= -x and passing through the origin.

THANK YOU :)

Sydney Sales's picture

DE: $(x²+4) y' + 3 xy = x$

(x²+4) y' + 3 xy = x

Differential Equations: $(r - 3s - 7) dr = (2r - 4s - 10) ds$

The topic is Additional Topics in Ordinary DE of the first order.

(r-3s-7)dr=(2r-4s-10)ds

Differential Equations: $(x - 2y - 1) dy = (2x - 4y - 5) dx$

Coefficient Linear in Two Variables? Help please.

(x-2y-1)dy=(2x-4y-5)dx

Sydney Sales's picture

Differential Equations: $[x \csc (y/x) - y] dx + x \, dy = 0$

2. (x csc y/x - y) dx + xdy=0
3. (x^2 + 2xy - 4y^2) dx - ( x^2 - 8xy - 4 y^2)=0
4. x^y ' = 4x^2 + 7xy + 2 y^2

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