The general solution of the differential equa | vedclass

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(C) Given equation: $(3y - 7x + 7)dx + (7y - 3x + 3)dy = 0$ Rearranging: $(7x - 3y - 7)dx + (3x - 7y - 3)dy = 0$ $\ldots(i)$ Solving the system $7x -

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$(x - y - 1)^2(x + y - 1)^5 = C$

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(C) Given equation: $(3y - 7x + 7)dx + (7y - 3x + 3)dy = 0$ Rearranging: $(7x - 3y - 7)dx + (3x - 7y - 3)dy = 0$ $\ldots(i)$ Solving the system $7x - 3y - 7 = 0$ and $3x - 7y - 3 = 0$,we get the intersection point $(1, 0)$. Substitute $x = 1 + u$ and $y = 0 + v = v$,so $dx = du$ and $dy = dv$. Substituting into $(i)$: $(7u - 3v)du + (3u - 7v)dv = 0$ $\ldots(ii)$ This is a homogeneous equation. Let $u = tv$,then $du = t dv + v dt$. Substituting into $(ii)$: $(7tv - 3v)(t dv + v dt) + (3tv - 7v)dv = 0$ $(7t - 3)(t dv + v dt) + (3t - 7)dv = 0$ $(7t^2 - 3t + 3t - 7)dv + v(7t - 3)dt = 0$ $(7t^2 - 7)dv + v(7t - 3)dt = 0$ $\int \frac{dv}{v} + \int \frac{7t - 3}{7(t^2 - 1)} dt = 0$ $\ln |v| + \frac{1}{7} \int \left( \frac{A}{t - 1} + \frac{B}{t + 1} \right) dt = C_1$ Using partial fractions: $\frac{7t - 3}{t^2 - 1} = \frac{2}{t - 1} + \frac{5}{t + 1}$. $\ln |v| + \frac{1}{7} [2 \ln |t - 1| + 5 \ln |t + 1|] = C_1$ $7 \ln |v| + 2 \ln |t - 1| + 5 \ln |t + 1| = C_2$ $\ln |v^7 (t - 1)^2 (t + 1)^5| = C_2$ Substituting $t = u/v$: $v^7 (u/v - 1)^2 (u/v + 1)^5 = C$ $v^7 \frac{(u - v)^2}{v^2} \frac{(u + v)^5}{v^5} = C$ $(u - v)^2 (u + v)^5 = C$ Substituting $u = x - 1$ and $v = y$: $(x - 1 - y)^2 (x - 1 + y)^5 = C$ $(x - y - 1)^2 (x + y - 1)^5 = C$

The general solution of the differential equation $(3y - 7x + 7)dx + (7y - 3x + 3)dy = 0$ is

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