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To solve the given differential equation, integrate both sides of the separated equation. We can do so as we integrate a of x with respect to x on the right side of the equation. Each integration requires a simple substitution. Let $u = 4y + 5$ and $v = 8x + 9$. Then $dy = \frac{du}{4}$ and $dx = \frac{dv}{8}$. $\int \frac{1}{(4y+5)^2} dy = \int \frac{1}{(8x+9)^2} dx$ $\int \frac{1}{u^2} \frac{1}{4} du = \frac{1}{8} \int \frac{1}{v^2} dv$ $-\frac{1}{4} u^{-1} + C_1 = \frac{1}{8} v^{-1} + C_2$ $-\frac{1}{4} (4y+5)^{-1} + C_1 = (\quad)(8x+9)^{-1} + C_2$ To finish, replace the constant $C_2 - C_1$ by $C$ to write the general solution to the given differential equation.

          To solve the given differential equation, integrate both sides of the separated equation. We can do so as we integrate a of x with respect to x on the right side of the equation.
Each integration requires a simple substitution. Let $u = 4y + 5$ and $v = 8x + 9$. Then $dy = \frac{du}{4}$ and $dx = \frac{dv}{8}$.
$\int \frac{1}{(4y+5)^2} dy = \int \frac{1}{(8x+9)^2} dx$
$\int \frac{1}{u^2} \frac{1}{4} du = \frac{1}{8} \int \frac{1}{v^2} dv$
$-\frac{1}{4} u^{-1} + C_1 = \frac{1}{8} v^{-1} + C_2$
$-\frac{1}{4} (4y+5)^{-1} + C_1 = (\quad)(8x+9)^{-1} + C_2$
To finish, replace the constant $C_2 - C_1$ by $C$ to write the general solution to the given differential equation.

to solve the given differential equation integrate both sides of the separated equation we can do so as we integrate a of x with respect to x on the right side of the equation each integrati 99433

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