The solution of the differential equation (2x | vedclass

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Question

(B) Given the differential equation: $(2x - 3y + 5)dx + (9y - 6x - 7)dy = 0$ Rearranging the terms: $\frac{dy}{dx} = -\frac{2x - 3y + 5}{-(6x - 9y + 7

Vedclass · Accepted Answer

$3x - 9y + 8 \log |6x - 9y - 1| = c$

Vedclass · Answer

(B) Given the differential equation: $(2x - 3y + 5)dx + (9y - 6x - 7)dy = 0$ Rearranging the terms: $\frac{dy}{dx} = -\frac{2x - 3y + 5}{-(6x - 9y + 7)} = \frac{2x - 3y + 5}{3(2x - 3y) + 7}$ Let $2x - 3y = z$. Then $2 - 3\frac{dy}{dx} = \frac{dz}{dx}$, which implies $\frac{dy}{dx} = \frac{1}{3}(2 - \frac{dz}{dx})$. Substituting into the equation: $\frac{1}{3}(2 - \frac{dz}{dx}) = \frac{z + 5}{3z + 7}$ $2 - \frac{dz}{dx} = \frac{3z + 15}{3z + 7}$ $\frac{dz}{dx} = 2 - \frac{3z + 15}{3z + 7} = \frac{6z + 14 - 3z - 15}{3z + 7} = \frac{3z - 1}{3z + 7}$ Separating variables: $\frac{3z + 7}{3z - 1} dz = dx$ Integrating both sides: $\int (1 + \frac{8}{3z - 1}) dz = \int dx$ $z + \frac{8}{3} \log |3z - 1| = x + c$ Multiply by $3$: $3z + 8 \log |3z - 1| = 3x + c'$ Substitute $z = 2x - 3y$: $3(2x - 3y) + 8 \log |3(2x - 3y) - 1| = 3x + c'$ $6x - 9y + 8 \log |6x - 9y - 1| = 3x + c'$ $3x - 9y + 8 \log |6x - 9y - 1| = c$.

The solution of the differential equation $(2x - 3y + 5)dx + (9y - 6x - 7)dy = 0$ is

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