(3.1) … … …
[tex]\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{2x-y}{x-2y}[/tex]
Multiply the right side by x/x :
[tex]\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{2-\dfrac yx}{1-\dfrac{2y}x}[/tex]
Substitute y(x) = x v(x), so that dy/dx = x dv/dx + v :
[tex]x\dfrac{\mathrm dv}{\mathrm dx} + v = \dfrac{2-v}{1-2v}[/tex]
This DE is now separable. With some simplification, you get
[tex]x\dfrac{\mathrm dv}{\mathrm dx} = \dfrac{2-2v+2v^2}{1-2v}[/tex]
[tex]\dfrac{1-2v}{2-2v+2v^2}\,\mathrm dv = \dfrac{\mathrm dx}x[/tex]
Now you're ready to integrate both sides (on the left, the denominator makes for a smooth substitution), which gives
[tex]-\dfrac12\ln\left|2v^2-2v+2\right| = \ln|x| + C[/tex]
Solve for v, then for y (or leave the solution in implicit form):
[tex]\ln\left|2v^2-2v+2\right| = -2\ln|x| + C[/tex]
[tex]\ln(2) + \ln\left|v^2-v+1\right| = \ln\left(\dfrac1{x^2}\right) + C[/tex]
[tex]\ln\left|v^2-v+1\right| = \ln\left(\dfrac1{x^2}\right) + C[/tex]
[tex]v^2-v+1 = e^{\ln\left(1/x^2\right)+C}[/tex]
[tex]v^2-v+1 = \dfrac C{x^2}[/tex]
[tex]\boxed{\left(\dfrac yx\right)^2 - \dfrac yx+1 = \dfrac C{x^2}}[/tex]
(3.2) … … …
[tex]y' + \dfrac yx = \dfrac{y^{-3/4}}{x^4}[/tex]
It may help to recognize this as a Bernoulli equation. Multiply both sides by [tex]y^{\frac34}[/tex] :
[tex]y^{3/4}y' + \dfrac{y^{7/4}}x = \dfrac1{x^4}[/tex]
Substitute [tex]z(x)=y(x)^{\frac74}[/tex], so that [tex]z' = \frac74 y^{3/4}y'[/tex]. Then you get a linear equation in z, which I write here in standard form:
[tex]\dfrac47 z' + \dfrac zx = \dfrac1{x^4} \implies z' + \dfrac7{4x}z=\dfrac7{4x^4}[/tex]
Multiply both sides by an integrating factor, [tex]x^{\frac74}[/tex], which gives
[tex]x^{7/4}z'+\dfrac74 x^{3/4}z = \dfrac74 x^{-9/4}[/tex]
and lets us condense the left side into the derivative of a product,
[tex]\left(x^{7/4}z\right)' = \dfrac74 x^{-9/4}[/tex]
Integrate both sides:
[tex]x^{7/4}z=\dfrac74\left(-\dfrac45\right) x^{-5/4}+C[/tex]
[tex]z=-\dfrac75 x^{-3} + Cx^{-7/4}[/tex]
Solve in terms of y :
[tex]y^{4/7}=-\dfrac7{5x^3} + \dfrac C{x^{7/4}}[/tex]
[tex]\boxed{y=\left(\dfrac C{x^{7/4}} - \dfrac7{5x^3}\right)^{7/4}}[/tex]
(3.3) … … …
[tex](\cos(x) - 2xy)\,\mathrm dx + \left(e^y-x^2\right)\,\mathrm dy = 0[/tex]
This DE is exact, since
[tex]\dfrac{\partial(-2xy)}{\partial y} = -2x[/tex]
[tex]\dfrac{\partial\left(e^y-x^2\right)}{\partial x} = -2x[/tex]
are the same. Then the general solution is a function f(x, y) = C, such that
[tex]\dfrac{\partial f}{\partial x}=\cos(x)-2xy[/tex]
[tex]\dfrac{\partial f}{\partial y} = e^y-x^2[/tex]
Integrating both sides of the first equation with respect to x gives
[tex]f(x,y) = \sin(x) - x^2y + g(y)[/tex]
Differentiating this result with respect to y then gives
[tex]-x^2 + \dfrac{\mathrm dg}{\mathrm dy} = e^y - x^2[/tex]
[tex]\implies\dfrac{\mathrm dg}{\mathrm dy} = e^y \implies g(y) = e^y + C[/tex]
Then the general solution is
[tex]\sin(x) - x^2y + e^y = C[/tex]
Given that y (1) = 4, we find
[tex]C = \sin(1) - 4 + e^4[/tex]
so that the particular solution is
[tex]\boxed{\sin(x) - x^2y + e^y = \sin(1) - 4 + e^4}[/tex]
