Answer to Question #169196 in Civil and Environmental Engineering for ezek

Question #169196

solve the follow homogenous equation of differential eq (2x-3y-1)dx (x 3y-5)dy=0


1
Expert's answer
2021-03-10T05:52:44-0500

We have given the differential equation,


(2x3y1)dx+(x3y5)dy=0(2x-3y-1)dx + (x -3y-5)dy = 0


dydx+(2x3y1)(x3y5)=0\dfrac{dy}{dx} + \dfrac{(2x -3y -1)}{(x -3y -5)} = 0


Putting x=X+hx = X+h and y=Y+ky = Y+k


2h3k1=0and h3k5=0\rightarrow 2h - 3k -1 = 0\\and\text{ } h - 3k -5 = 0


Hence, dYdX+(2X3Y)(X+3Y)=0\dfrac{dY}{dX} + \dfrac{(2X-3Y)}{(X+3Y)} = 0


Putting Y=vXY = vX


v+XdvdX+23v13v=0v+X\dfrac{dv}{dX} + \dfrac{2-3v}{1-3v} = 0


XdvdX=(3v2)(13v)vX\dfrac{dv}{dX} = \dfrac{(3v-2)}{(1-3v)} -v


=(2v23v2)(13v)= \dfrac{(2v-2-3v^2)}{(1-3v)}


(13v2v23v2)dv+dXX=logC\int(\dfrac{1-3v}{2v-2-3v^2})dv + \int\dfrac{dX}{X} = logC


12log(2v23v2)+logX=logc\dfrac{1}{2}log(2v-2-3v^2)+logX=logc


Hence, this is the final solution of the given differential equation.




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