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Solve the D.E:
(dy)/(dx)=(x+y+1)/(2x+2y+)

Solve the D.E: $\frac{d y}{d x}=\frac{x+y+1}{2 x+2 y+}$

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Sum of finite series not start from 1

Full solution

Q. Solve the D.E:

\frac{d y}{d x}=\frac{x+y+1}{2 x+2 y+}

Rewrite equation: Rewrite the differential equation in a more simplified form by dividing both sides by $2$ . $\frac{dy}{dx} = \frac{\frac{x}{2} + \frac{y}{2} + \frac{1}{2}}{x + y}$
Integrating factor: Notice that the equation is not separable as is, so we look for an integrating factor or a substitution that can simplify it. Let's try the substitution $u = x + y$ .
Find $\frac{du}{dx}$ : Differentiate $u$ with respect to $x$ to find $\frac{du}{dx}$ .
$\frac{du}{dx} = \frac{dx}{dx} + \frac{dy}{dx}$
$\frac{du}{dx} = 1 + \frac{dy}{dx}$
Substitute $\frac{du}{dx}$ : Substitute $\frac{dy}{dx}$ from the original equation into the $\frac{du}{dx}$ equation. $\newline$ $\frac{du}{dx} = 1 + \frac{x + y + 1}{2x + 2y}$
Rewrite with $u$ : Since $u = x + y$ , we can rewrite the equation as: $\newline$ $\frac{du}{dx} = 1 + \frac{u + 1}{2u}$
Simplify equation: Simplify the equation. $\newline$ $\frac{du}{dx} = \frac{2u + u + 1}{2u}$ $\newline$ $\frac{du}{dx} = \frac{3u + 1}{2u}$
Separate variables: Separate variables to integrate. $\newline$ $\frac{2u}{3u + 1} \, du = dx$
Integrate both sides: Integrate both sides. $\newline$ $\int\frac{2u}{3u + 1}\,du = \int dx$
Partial fraction decomposition: Use partial fraction decomposition on the left side. $\newline$ Let $\frac{2u}{3u + 1} = \frac{A}{3u + 1}$ , then $2u = A$ . $\newline$ $A = 2$ .
Integrate with $A$ : Integrate using the determined $A$ value. $\newline$ $\int\frac{2}{3u + 1} du = \int dx$
Perform integration: Perform the integration. $(\frac{2}{3})\ln|3u + 1| = x + C$
Solve for $u$ : Solve for $u$ by exponentiating both sides. $\newline$ |\(3u + $1$ | = e^{\frac{ $3$ }{ $2$ }(x + C)}|
Exponentiate both sides: Remove the absolute value by considering both positive and negative cases. $\newline$ $3u + 1 = \pm e^{\frac{3}{2}(x + C)}$
Remove absolute value: Solve for $u$ . $u = \frac{\pm e^{\frac{3}{2}(x + C)} - 1}{3}$
Solve for $u$ : Substitute back $u = x + y$ . $x + y = (\pm e^{\frac{3}{2}(x + C)} - 1)/3$
Substitute back $u$ : Solve for $y$ . $y = \left(\pm e^{\frac{3}{2}(x + C)} - 1\right)/3 - x$

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