The general solution of the differential equa | vedclass

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(C) Given differential equation: $(x-2y+1)dy - (3x-6y+2)dx = 0$ $\Rightarrow \frac{dy}{dx} = \frac{3x-6y+2}{x-2y+1} = \frac{3(x-2y)+2}{(x-2y)+1}$ Let

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$\left|x-2y+\frac{3}{5}\right|^{2/25} \cdot e^{1/5(6x-2y)}=C$

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(C) Given differential equation: $(x-2y+1)dy - (3x-6y+2)dx = 0$ $\Rightarrow \frac{dy}{dx} = \frac{3x-6y+2}{x-2y+1} = \frac{3(x-2y)+2}{(x-2y)+1}$ Let $v = x-2y$. Then $\frac{dv}{dx} = 1 - 2\frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{1}{2}(1 - \frac{dv}{dx})$. Substituting this into the equation: $\frac{1}{2}(1 - \frac{dv}{dx}) = \frac{3v+2}{v+1}$ $1 - \frac{dv}{dx} = \frac{6v+4}{v+1} \Rightarrow \frac{dv}{dx} = 1 - \frac{6v+4}{v+1} = \frac{v+1-6v-4}{v+1} = \frac{-5v-3}{v+1}$ $\int \frac{v+1}{-5v-3} dv = \int dx$ $-\frac{1}{5} \int \frac{5v+5}{5v+3} dv = x + C_1$ $-\frac{1}{5} \int \frac{(5v+3)+2}{5v+3} dv = x + C_1$ $-\frac{1}{5} [v + \frac{2}{5} \ln|5v+3|] = x + C_1$ Substitute $v = x-2y$: $-\frac{1}{5}(x-2y) - \frac{2}{25} \ln|5(x-2y)+3| = x + C_1$ $-\frac{2}{25} \ln|5x-10y+3| = x + \frac{1}{5}x - \frac{2}{5}y + C_1 = \frac{6}{5}x - \frac{2}{5}y + C_1$ $\ln|5x-10y+3|^{-2/25} = \frac{6x-2y}{5} + C_1$ $|5x-10y+3|^{-2/25} = e^{\frac{6x-2y}{5}} \cdot e^{C_1}$ $|5(x-2y+\frac{3}{5})|^{-2/25} = e^{\frac{6x-2y}{5}} \cdot C_2$ $|x-2y+\frac{3}{5}|^{-2/25} = e^{\frac{6x-2y}{5}} \cdot C_3$ $\left|x-2y+\frac{3}{5}\right|^{2/25} \cdot e^{\frac{1}{5}(6x-2y)} = C$

The general solution of the differential equation $(x-2y+1)dy-(3x-6y+2)dx=0$ is

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