Question

(x+y-4)dx-(3x-y-4)dy=0 when x=4 and y=7.

Answer 1

(x+y-4)dx-(3x-y-4)dy=0 

$\frac{dy}{dx} = \frac{x+y-4}{3x-y-4}$

Let x = u+h and. y = v+k

$\frac{x+y-4}{3x-y-4}$ = $\frac{u+h+v+k-4}{3u+3h-v-k-4}=\frac{u+v+h+k-4}{3u-v+3h-k-4}$

To make it homogeneous put

h+k-4=0 and 3h-k-4=0

Adding we get 4h = 8=> h = 2

So k = 2

So x = u+2 and y = v+2

dx = du and dy = dv

So $\frac{dy}{dx} = \frac{dv}{du}$

Now the given differential equation transforms to

$\frac{dv}{du} = \frac{u+v}{3u-v}$

Let v = tu

$\frac{dv}{du} = t + u \frac{dt}{du}$

=> $t + u \frac{dt}{du} =$ $\frac{u+tu}{3u-tu}$

=> $u \frac{dt}{du} =$ $\frac{1+t}{3-t}-t$

=> $u \frac{dt}{du} =$ $\frac{1-2t+t²}{3-t}$

=> $\frac{3-t}{(t-1)²} dt = \frac{du}{u}$

=> $\frac{2-(t-1)}{(t-1)²} dt = \frac{du}{u}$

=> $\frac{2}{(t-1)²} dt -\frac{1}{(t-1)} dt=\frac{du}{u}$

=> $\int\frac{2}{(t-1)²} dt -\int\frac{1}{(t-1)} dt=\int\frac{du}{u}$

=> $-\frac{2}{t-1} - ln|t-1|=ln|u| - lnC$

=> $-\frac{2}{t-1} = ln|\frac{u(t-1)}{C}|$

=> $-\frac{2u}{v-u} = ln|\frac{(v-u)}{C}|$

=> $-\frac{2(x-2)}{y-x} = ln|\frac{(y-x)}{C}|$

=> $\frac{(y-x)}{C}=$ $e^{-\frac{2(x-2)}{y-x} }$

=> y = x + $Ce^{-\frac{2(x-2)}{y-x} }$ , This is the general solution.

Now y = 7 when x = 4

So 7 = 4 + $Ce^{-\frac{2(4-2)}{7-4} }$

=> C = 3$e^{\frac{4}{3} }$

So the solution is

y = x $+e^{\frac{4}{3}}.e^{-\frac{2(x-2)}{y-x} }$