The particular solution of the differential e | vedclass

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Question

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

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$2x - y - \log(x - y + 2) + 1 = 0$

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(D) Given differential equation: $(2x - 2y + 3)dx - (x - y + 1)dy = 0$ $\Rightarrow \frac{dy}{dx} = \frac{2(x - y) + 3}{x - y + 1}$ Let $v = x - y$. Then $\frac{dv}{dx} = 1 - \frac{dy}{dx}$,so $\frac{dy}{dx} = 1 - \frac{dv}{dx}$. Substituting into the equation: $1 - \frac{dv}{dx} = \frac{2v + 3}{v + 1}$ $\Rightarrow \frac{dv}{dx} = 1 - \frac{2v + 3}{v + 1} = \frac{v + 1 - 2v - 3}{v + 1} = \frac{-(v + 2)}{v + 1}$ $\Rightarrow \frac{v + 1}{v + 2} dv = -dx$ $\Rightarrow \int \left(1 - \frac{1}{v + 2}\right) dv = \int -dx$ $\Rightarrow v - \log|v + 2| = -x + C$ Substitute $v = x - y$: $(x - y) - \log|x - y + 2| = -x + C$ $\Rightarrow 2x - y - \log|x - y + 2| = C$ Given $x = 0, y = 1$: $2(0) - 1 - \log|0 - 1 + 2| = C \Rightarrow -1 - \log(1) = C \Rightarrow C = -1$ Thus,the particular solution is $2x - y - \log(x - y + 2) = -1$ or $2x - y - \log(x - y + 2) + 1 = 0$.

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

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