Question

Solve the following

dy/dx=(1-3y-3x) /(1+x+y)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Differential equation is

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1 - 3y - 3x}{1 + x + y}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1 - 3(y + x)}{1 + x + y}$

Now, we Substitute

$\red{\rm :\longmapsto\:y + x = z}$

$\rm :\longmapsto\:\dfrac{d}{dx}y + \dfrac{d}{dx}x = \dfrac{d}{dx}z$

$\rm :\longmapsto\:\dfrac{dy}{dx} + 1 = \dfrac{dz}{dx}$

$\red{\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{dz}{dx} - 1}$

On substituting all these values in above differential equation, we get

$\rm :\longmapsto\: \dfrac{dz}{dx} - 1 = \dfrac{1 - 3z}{1 + z}$

$\rm :\longmapsto\: \dfrac{dz}{dx} = \dfrac{1 - 3z}{1 + z} + 1$

$\rm :\longmapsto\: \dfrac{dz}{dx} = \dfrac{1 - 3z + 1 + z}{1 + z}$

$\rm :\longmapsto\: \dfrac{dz}{dx} = \dfrac{2 - 2z}{1 + z}$

$\rm :\longmapsto\: \dfrac{dz}{dx} = \dfrac{2(1 - z)}{1 + z}$

On separating the variables, we get

$\rm :\longmapsto\:\dfrac{z + 1}{1 - z}dz = 2dx$

On integrating both sides, we get

$\rm :\longmapsto\:\displaystyle\int \dfrac{z + 1}{1 - z}dz = \displaystyle\int 2dx$

$\rm :\longmapsto\: - \displaystyle\int \dfrac{z + 1}{z - 1}dz = 2x + c$

$\rm :\longmapsto\: - \displaystyle\int \dfrac{z - 1 + 2}{z - 1}dz = 2x + c$

$\rm :\longmapsto\: - \displaystyle\int\bigg[1 + \dfrac{2}{z - 1}\bigg]dz = 2x + c$

$\rm :\longmapsto\: - (z + log |z - 1|) = 2x + c$

$\rm :\longmapsto\: - z - log |z - 1| = 2x + c$

$\rm :\longmapsto\: - (x + y) - log |x + y - 1| = 2x + c$

$\rm :\longmapsto\: - x - y - log |x + y - 1| = 2x + c$

$\rm :\longmapsto\: 3x + y + log |x + y - 1| = - c$

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Formula Used

$\boxed{\tt{ \displaystyle\int {x}^{n} = \frac{ {x}^{n + 1} }{n + 1} + c}}$

$\boxed{\tt{ \displaystyle\int \frac{1}{x} = log |x| + c}}$

$\boxed{\tt{ \displaystyle\int kdx = kx + c}}$

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EXPLORE MORE

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$