Question

Find the General Solution of the differential Equation.

Non - Homogeneous Differential Equation.

Using Integration

Question = (2x + y + 1)dx = (4x + 2y - 1)dy = 0

Answer = x + 2y + log(2x + y - 1) = c

Answer 1

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$= > (2x + y + 1)dx + (4x + 2y - 1)dy = 0 \\ \\ = > (2x + y + 1)dx = - (4x + 2y - 1)dy \\ \\ = > \frac{dy}{dx} = - \frac{(2x + y) + 1}{2(2x + y) - 1} \\ \\ \bf \: let \: 2x + y = z \\ \\ = > \frac{dy}{dx} = \frac{dz}{dx} - 2 \\ \\ \bf \: \: \: \: \: \: \: \: \underline{now} \\ \\ = > \frac{dz}{dx} - 2 = - \frac{z + 1}{2z - 1 } \\ \\ = > \frac{dz}{dx} = - \frac{z + 1}{2z - 1} + 2 = \frac{3z - 3}{2z - 1} \\ \\ = > \frac{2z - 1}{z - 1} dz = 3dx \\ \\ = > \int \: (2 + \frac{1}{z - 1} )dz = 3 \int \: dx \\ \\ = > 2 \int \: dz + \int \frac{1}{z - 1} dz = 3 \int \: dx \\ \\ = > 2z + log(z - 1) = 3x + c \\ \\ = > 2(2x + y) + log(2x + y - 1) = 3x + c \\ \\ = > 4x - 3x + 2y + log(2x + y - 1) = c \\ \\ = > \bf \: x + 2y + log(2x + y - 1) = c$
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