The general solution of the differential equa | vedclass

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Question

(C) Given the differential equation $\frac{d y}{d x}=\frac{x+7 y+3}{3 x+5 y+9}$.Let $x = X+h$ and $y = Y+k$. Then $\frac{d y}{d x} = \frac{d Y}{d X}$.

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$(y-x+3)^4=c|5 y+x-3|$

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(C) Given the differential equation $\frac{d y}{d x}=\frac{x+7 y+3}{3 x+5 y+9}$.Let $x = X+h$ and $y = Y+k$. Then $\frac{d y}{d x} = \frac{d Y}{d X}$.Substituting these into the equation, we get $\frac{d Y}{d X} = \frac{X+7 Y + (h+7k+3)}{3X+5Y + (3h+5k+9)}$.To make this homogeneous, we set $h+7k+3=0$ and $3h+5k+9=0$.Solving these linear equations, we find $h=-3$ and $k=0$.Thus, the equation becomes $\frac{d Y}{d X} = \frac{X+7 Y}{3X+5Y}$.Let $Y = vX$, then $\frac{d Y}{d X} = v + X \frac{d v}{d X}$.$v + X \frac{d v}{d X} = \frac{1+7v}{3+5v} \Rightarrow X \frac{d v}{d X} = \frac{1+7v - 3v - 5v^2}{3+5v} = \frac{-5v^2+4v+1}{3+5v}$.Separating variables: $\int \frac{5v+3}{5v^2-4v-1} dv = -\int \frac{1}{X} dX$.Using partial fractions or integration techniques, we solve the integral to obtain the relation between $v$ and $X$.Substituting back $v = \frac{Y}{X} = \frac{y}{x+3}$ and $X = x+3$, we arrive at the general solution $(y-x+3)^4 = c|5y+x-3|$.

The general solution of the differential equation $\frac{d y}{d x}=\frac{x+7 y+3}{3 x+5 y+9}$ is

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