Question

Find the General Solution of the differential Equation.

Non - Homogeneous Differential Equation.

Using Integration

Question = dy/dx = x + y - 1/2x + 2y + 3

Answer = [ 6y - 3x + log(3x + 3y + 4)] = 0

Answer 1

Hope u like my process
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$= > \frac{dy}{dx} = \frac{x + y + 1}{2x + 2y + 3} \\ \\ = > \frac{dy}{dx} = \frac{2x + 2y + 2}{2(2x + 2y + 3)} \\ \\ = > 2 \frac{dy}{dx} = \frac{(2x + 2y + 3) - 1}{(2x + 2y + 3)} \\ \\ \bf \: let \: \: 2x + 2y + 3= z \\ \\ = > 2\frac{dy}{dx} = \frac{dz}{dx} - 2\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \underline{now} \\ \\ = > \frac{dz}{dx} - 2 = \frac{z - 1}{z} \\ \\ = > \frac{dz}{dx} = 2 + \frac{z - 1}{z} = \frac{3z - 1}{z} \\ \\ = > \frac{z}{z - \frac{1}{3} } dz = 3dx \\ \\ = > \int(1 + \frac{ \frac{1}{3} }{z - \frac{1}{3} } )dz = 3 \int \: dx \\ \\ = > \int dz + \frac{1}{3} \int \frac{1}{z - \frac{1}{3} } dz = 3 \int \: dx \\ \\ = > z + \frac{1}{3} log(z - \frac{1}{3} ) = 3x + c \\ \\ = > 3z + log(3z - 1) - log(3) = 9x + c \\ \\ = > 3(2x + 2y + 3) + log(3(2x + 2y + 3) - 1) - 9x = c \\ \\ = > 6x + 6y + 9 + log(6x + 6y + 9 - 1) - 9x = c \\ \\ = > 6y - 3x + log(3x + 3y + 4) + log(2) = c \\ \\ \bf = > 6y - 3x + log(3x + 3y + 4) = c$
this is the required general solution
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