Question

"Find the general solution of diffirential equation (x-y-2)dx-(2x-2y-3)dy=0"

Answer 1

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

Given Differential equation is

$\rm \: (x - y - 2)dx - (2x - 2y - 3)dy = 0$

can be rewritten as

$\rm \: (x - y - 2)dx = (2x - 2y - 3)dy$

$\rm \: \dfrac{dy}{dx} = \dfrac{x - y - 2}{2x - 2y - 3}$

$\rm \: \dfrac{dy}{dx} = \dfrac{x - y - 2}{2(x - y) - 3}$

Now, Substitute

$\rm \: x - y = z$

On differentiating both sides w. r. t. x, we get

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

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

So, Substitute all these values in above expression, we get

$\rm \: 1 - \dfrac{dz}{dx} = \dfrac{z - 2}{2z - 3}$

$\rm \: \dfrac{dz}{dx} = 1 - \dfrac{z - 2}{2z - 3}$

$\rm \: \dfrac{dz}{dx} = \dfrac{2z - 3 - z + 2}{2z - 3}$

$\rm \: \dfrac{dz}{dx} = \dfrac{z - 1}{2z - 3}$

On separating the variables, we get

$\rm \: \dfrac{2z - 3}{z - 1} \: dz = dx$

On integrating both sides, we get

$\rm \: \displaystyle\int\rm \dfrac{2z - 3}{z - 1} \: dz =\displaystyle\int\rm dx$

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

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

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

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

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

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

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

So, required solution is

$\rm\implies \:\boxed{ \rm{ \:\rm \: x - 2y- log |x - y - 1| = c \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

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

Answer 2

Step-by-step explanation:

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

Solution−

Given Differential equation is

\begin{gathered}\rm \: (x - y - 2)dx - (2x - 2y - 3)dy = 0 \\ \end{gathered}

(x−y−2)dx−(2x−2y−3)dy=0

can be rewritten as

\begin{gathered}\rm \: (x - y - 2)dx = (2x - 2y - 3)dy \\ \end{gathered}

(x−y−2)dx=(2x−2y−3)dy

\begin{gathered}\rm \: \dfrac{dy}{dx} = \dfrac{x - y - 2}{2x - 2y - 3} \\ \end{gathered}

dx

dy

=

2x−2y−3

x−y−2

\begin{gathered}\rm \: \dfrac{dy}{dx} = \dfrac{x - y - 2}{2(x - y) - 3} \\ \end{gathered}

dx

dy

=

2(x−y)−3

x−y−2

Now, Substitute

\begin{gathered}\rm \: x - y = z \\ \end{gathered}

x−y=z

On differentiating both sides w. r. t. x, we get

\begin{gathered}\rm \: 1 - \dfrac{dy}{dx} = \dfrac{dz}{dx} \\ \end{gathered}

1−

dx

dy

=

dx

dz

\begin{gathered}\rm\implies \:\rm \: \dfrac{dy}{dx} =1 - \dfrac{dz}{dx} \\ \end{gathered}

⟹

dx

dy

=1−

dx

dz

So, Substitute all these values in above expression, we get

\begin{gathered}\rm \: 1 - \dfrac{dz}{dx} = \dfrac{z - 2}{2z - 3} \\ \end{gathered}

1−

dx

dz

=

2z−3

z−2

\begin{gathered}\rm \: \dfrac{dz}{dx} = 1 - \dfrac{z - 2}{2z - 3} \\ \end{gathered}

dx

dz

=1−

2z−3

z−2

\begin{gathered}\rm \: \dfrac{dz}{dx} = \dfrac{2z - 3 - z + 2}{2z - 3} \\ \end{gathered}

dx

dz

=

2z−3

2z−3−z+2

\begin{gathered}\rm \: \dfrac{dz}{dx} = \dfrac{z - 1}{2z - 3} \\ \end{gathered}

dx

dz

=

2z−3

z−1

On separating the variables, we get

\begin{gathered}\rm \: \dfrac{2z - 3}{z - 1} \: dz = dx \\ \end{gathered}

z−1

2z−3

dz=dx

On integrating both sides, we get

\begin{gathered}\rm \: \displaystyle\int\rm \dfrac{2z - 3}{z - 1} \: dz =\displaystyle\int\rm dx \\ \end{gathered}

∫

z−1

2z−3

dz=∫dx

\begin{gathered}\rm \: \displaystyle\int\rm \dfrac{2z - 2 - 1}{z - 1} \: dz = x + c \\ \end{gathered}

∫

z−1

2z−2−1

dz=x+c

\begin{gathered}\rm \: \displaystyle\int\rm \dfrac{2(z - 1) - 1}{z - 1} \: dz = x + c \\ \end{gathered}

∫

z−1

2(z−1)−1

dz=x+c

\begin{gathered}\rm \: \displaystyle\int\rm \bigg(2 - \dfrac{1}{z - 1} \bigg) \: dz = x + c \\ \end{gathered}

∫(2−

z−1

1

)dz=x+c

\begin{gathered}\rm \: 2z - log |z - 1| = x + c \\ \end{gathered}

2z−log∣z−1∣=x+c

\begin{gathered}\rm \: 2(x - y) - log |x - y - 1| = x + c \\ \end{gathered}

2(x−y)−log∣x−y−1∣=x+c

\begin{gathered}\rm \: 2x - 2y- log |x - y - 1| = x + c \\ \end{gathered}

2x−2y−log∣x−y−1∣=x+c

\begin{gathered}\rm \: x - 2y- log |x - y - 1| = c \\ \end{gathered}

x−2y−log∣x−y−1∣=c

So, required solution is

\begin{gathered}\rm\implies \:\boxed{ \rm{ \:\rm \: x - 2y- log |x - y - 1| = c \: \: }} \\ \end{gathered}

⟹

x−2y−log∣x−y−1∣=c

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\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}\end{gathered}

f(x)

k

sinx

cosx

sec

2

x

cosec

2

x

secxtanx

cosecxcotx

tanx

x

1

e

x

∫f(x)dx

kx+c

−cosx+c

sinx+c

tanx+c

−cotx+c

secx+c

−cosecx+c

logsecx+c

logx+c

e

x

+c