Classify the following differential equations as separable, homogeneous, parallel line,

Classify the following differential equations as separable, homogeneous, parallel line, or exact. Explain briefly your answers. Then, solve each equation according to their classification. $$\displaystyle{\left({2}{x}-{3}{y}\right)}{\left.{d}{x}\right.}+{\left({2}{y}-{3}{x}\right)}{\left.{d}{y}\right.}={0}$$

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Archie Jones
Classsify the following differential eq.
$$\displaystyle{\left({2}{x}-{3}{y}\right)}{\left.{d}{x}\right.}+{\left({2}{y}-{3}{x}\right)}{\left.{d}{y}\right.}={0}$$
Solution:
$$\displaystyle{\left.{d}{y}\right.}{\left({2}{y}-{3}{x}\right)}=-{\left({2}{x}-{3}{y}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}=-{\frac{{{\left({2}{x}-{3}{y}\right)}}}{{{\left({2}{y}-{3}{x}\right)}}}}$$
Finding homogeneous
$$\displaystyle{F}{\left(\lambda{x},\lambda{y}\right)}={\frac{{-{\left({2}\lambda{x}-{3}\lambda{y}\right)}}}{{{\left({2}\lambda{y}-{3}\lambda{x}\right)}}}}$$
$$\displaystyle={\frac{{-\lambda{\left({2}{x}-{3}{y}\right)}}}{{{\left({2}{y}-{3}{x}\right)}}}}$$
$$\displaystyle={\frac{{-{\left({2}{x}-{3}{y}\right)}}}{{{\left({2}{x}-{3}{x}\right)}}}}$$
$$\displaystyle=\lambda^{{\circ}}{F}{\left({r},{y}\right)}$$
The given equation is homogenous.
Not exactly what you’re looking for?
Robert Pina
Simplifying
$$\displaystyle{\left({2}{x}+-{3}{y}\right)}\cdot{\left.{d}{x}\right.}+{\left({2}{y}+-{3}{x}\right)}\cdot{\left.{d}{y}\right.}={0}$$
Reorder the terms for easier multiplication:
$$\displaystyle{\left.{d}{x}\right.}{\left({2}{x}+-{3}{y}\right)}+{\left({2}{y}+-{3}{x}\right)}\cdot{\left.{d}{y}\right.}={0}$$
$$\displaystyle{\left({2}{x}\cdot{\left.{d}{x}\right.}+-{3}{y}\cdot{\left.{d}{x}\right.}\right)}+{\left({2}{y}+-{3}{x}\right)}\cdot{\left.{d}{y}\right.}={0}$$
Reorder the terms:
$$\displaystyle{\left(-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}\right)}+{\left({2}{y}+-{3}{x}\right)}\cdot{\left.{d}{y}\right.}={0}$$
$$\displaystyle{\left(-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}\right)}+{\left({2}{y}+-{3}{x}\right)}\cdot{\left.{d}{y}\right.}={0}$$
Reorder the terms:
$$\displaystyle-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}+{\left(-{3}{x}+{2}{y}\right)}\cdot{\left.{d}{y}\right.}={0}$$
Reorder the terms for easier multiplication:
$$\displaystyle-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}+{\left.{d}{y}\right.}{\left(-{3}{x}+{2}{y}\right)}={0}$$
$$\displaystyle-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}+{\left(-{3}{x}\cdot{\left.{d}{y}\right.}+{2}{y}\cdot{\left.{d}{y}\right.}\right)}={0}$$
$$\displaystyle-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}+{\left(-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{y}\right.}^{{{2}}}\right)}={0}$$
Reorder the terms:
$$\displaystyle-{3}{\left.{d}{x}\right.}{y}+-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}+{2}{\left.{d}{y}\right.}^{{{2}}}={0}$$
Combine like terms: $$\displaystyle-{3}{\left.{d}{x}\right.}{y}+-{3}{\left.{d}{x}\right.}{y}=-{6}{\left.{d}{x}\right.}{y}$$
$$\displaystyle-{6}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}+{2}{\left.{d}{y}\right.}^{{{2}}}={0}$$
Solving
$$\displaystyle-{6}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}+{2}{\left.{d}{y}\right.}^{{{2}}}={0}$$
Solving for variable 'd'.
Move all terms containing d to the left, all other terms to the right.
Factor out the Greatest Common Factor (GCF), '2d'.
$$\displaystyle{2}{d}{\left(-{3}{x}{y}+{x}^{{{2}}}+{y}^{{{2}}}\right)}={0}$$
karton

$$2x-3y+(2y-3x) \frac{dy}{dx}=0$$
$$2x-3y+(2y-3x)y'=0$$
Verify that $$\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}$$: True
Find $$(x,y):(x,y)=y^{2}-3xy+x^{2}+c_{1}$$
$$y^{2}-3xy+x^{2}+c_{1}=c_{2}$$
$$y^{2}-3xy+x^{2}=c_{1}$$
Isolate y: $$y=\frac{3x+\sqrt{5x^{2}+4c_{1}}}{2}, y=\frac{3x-\sqrt{5x^{2}+4c_{1}}}{2}$$
$$y=\frac{3x+\sqrt{5x^{2}+c_{1}}}{2}, y=\frac{3x-\sqrt{5x^{2}+c_{1}}}{2}$$