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DivGradCurl

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Problem from my textbook - Solve:

[tex]\frac{dy}{dx} = \frac{4y - 3x}{2x - y}[/tex]

Answer provided:

[tex]\left| y - x \right| = \mathrm{C} \left| y + 3x \right| ^5 \mbox{.} \mbox{ Also, } y = -3x \mbox{.}[/tex]

My answer is pretty close to that one, but it seems like I missed some step. Anyhow, here's what's I've got:

Left-hand side:

[tex] \frac{dy}{dx} = \frac{d}{dx} \left( xv \right)[/tex]

[tex]\frac{dy}{dx} = xv^{\prime} + v[/tex]

[tex]\frac{dy}{dx} = x\frac{dv}{dx} + v[/tex]

Right-hand side:

[tex]\frac{dy}{dx} = \frac{4y - 3(y/v)}{2(y/v) - y}[/tex]

[tex] \frac{dy}{dx} = -\left( \frac{4v-3}{v-2} \right)[/tex]

Hence, this equation is homogeneous. As a result, we find

[tex] x\frac{dv}{dx} + v = -\left( \frac{4v-3}{v-2} \right) [/tex]

[tex]x\frac{dv}{dx} = \frac{-v^2 -2v + 3}{v-2} \mbox{,}[/tex]

which is separable. Thus, we obtain

[tex]\int \frac{v-2}{-v^2 -2v + 3} \: dv = \int \frac{dx}{x}[/tex]

[tex]\frac{1}{4} \ln \left| v - 1 \right| - \frac{5}{4} \ln \left| v + 3 \right| = \ln \left| x \right| + \mathrm{C}[/tex]

[tex]\ln \left| v - 1 \right| - 5 \ln \left| v + 3 \right| = 4 \ln \left| x \right| + \mathrm{C}[/tex]

[tex]\ln \frac{\left| v - 1 \right|}{\left| v + 3 \right| ^5} = \ln \left( x^4 \right) + \mathrm{C}[/tex]

[tex]\frac{\left| v - 1 \right|}{\left| v + 3 \right| ^5} = \mathrm{C} x^4[/tex]

[tex]\frac{\left| (y/x) - 1 \right|}{\left| (y/x) + 3 \right| ^5} = \mathrm{C} x^4[/tex]

[tex]\frac{\left| y - x \right|}{\left| y + 3x \right| ^5} = \mathrm{C} x^4[/tex]

[tex]\left| y - x \right| = \mathrm{C} x^4 \left| y + 3x \right| ^5 \mbox{,}[/tex]

which is different from

[tex]\left| y - x \right| = \mathrm{C} \left| y + 3x \right| ^5 \mbox{.}[/tex]

Again, it seems like I missed some step. Can anyone help me out with that?

By the way, I can't see why [tex]y=-3x[/tex] also is an answer. Could you please clarify it?

Any help is highly appreciated.

[tex]\frac{dy}{dx} = \frac{4y - 3x}{2x - y}[/tex]

Answer provided:

[tex]\left| y - x \right| = \mathrm{C} \left| y + 3x \right| ^5 \mbox{.} \mbox{ Also, } y = -3x \mbox{.}[/tex]

My answer is pretty close to that one, but it seems like I missed some step. Anyhow, here's what's I've got:

Left-hand side:

[tex] \frac{dy}{dx} = \frac{d}{dx} \left( xv \right)[/tex]

[tex]\frac{dy}{dx} = xv^{\prime} + v[/tex]

[tex]\frac{dy}{dx} = x\frac{dv}{dx} + v[/tex]

Right-hand side:

[tex]\frac{dy}{dx} = \frac{4y - 3(y/v)}{2(y/v) - y}[/tex]

[tex] \frac{dy}{dx} = -\left( \frac{4v-3}{v-2} \right)[/tex]

Hence, this equation is homogeneous. As a result, we find

[tex] x\frac{dv}{dx} + v = -\left( \frac{4v-3}{v-2} \right) [/tex]

[tex]x\frac{dv}{dx} = \frac{-v^2 -2v + 3}{v-2} \mbox{,}[/tex]

which is separable. Thus, we obtain

[tex]\int \frac{v-2}{-v^2 -2v + 3} \: dv = \int \frac{dx}{x}[/tex]

[tex]\frac{1}{4} \ln \left| v - 1 \right| - \frac{5}{4} \ln \left| v + 3 \right| = \ln \left| x \right| + \mathrm{C}[/tex]

[tex]\ln \left| v - 1 \right| - 5 \ln \left| v + 3 \right| = 4 \ln \left| x \right| + \mathrm{C}[/tex]

[tex]\ln \frac{\left| v - 1 \right|}{\left| v + 3 \right| ^5} = \ln \left( x^4 \right) + \mathrm{C}[/tex]

[tex]\frac{\left| v - 1 \right|}{\left| v + 3 \right| ^5} = \mathrm{C} x^4[/tex]

[tex]\frac{\left| (y/x) - 1 \right|}{\left| (y/x) + 3 \right| ^5} = \mathrm{C} x^4[/tex]

[tex]\frac{\left| y - x \right|}{\left| y + 3x \right| ^5} = \mathrm{C} x^4[/tex]

[tex]\left| y - x \right| = \mathrm{C} x^4 \left| y + 3x \right| ^5 \mbox{,}[/tex]

which is different from

[tex]\left| y - x \right| = \mathrm{C} \left| y + 3x \right| ^5 \mbox{.}[/tex]

Again, it seems like I missed some step. Can anyone help me out with that?

By the way, I can't see why [tex]y=-3x[/tex] also is an answer. Could you please clarify it?

Any help is highly appreciated.

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