Question #115575
find the general integral of the equation (x-y)p+(y-x-z)q=z and a particular solution through the circle z=1,x^2+y^2=1
1
Expert's answer
2020-05-19T08:43:02-0400

By Lagrange’s method the auxiliary equations of the given differential equation are as following:

𝑑𝑥𝑥𝑦=𝑑𝑦𝑦𝑥𝑧=𝑑𝑧𝑧\frac{𝑑𝑥}{𝑥−𝑦}=\frac{𝑑𝑦}{𝑦−𝑥−𝑧}=\frac{𝑑𝑧}{𝑧}

    𝑑𝑥𝑥𝑦=𝑑𝑦𝑦𝑥𝑧=𝑑𝑧𝑧=dx+dy+dz0    dx+dy+dz=0    x+y+z=c1\implies \frac{𝑑𝑥}{𝑥−𝑦}=\frac{𝑑𝑦}{𝑦−𝑥−𝑧}=\frac{𝑑𝑧}{𝑧} = \frac{dx+dy+dz}{0} \\ \implies dx+dy+dz =0 \implies x+y+z = c_1

Hence from the first fraction, we have 𝑑𝑥𝑥𝑦=𝑑𝑦𝑦𝑥𝑧\frac{𝑑𝑥}{𝑥−𝑦}=\frac{𝑑𝑦}{𝑦−𝑥−𝑧}

    𝑑𝑦𝑦c1+y=𝑑zz\implies \frac{𝑑𝑦}{𝑦−c_1+y}=\frac{𝑑z}{z}

Now, by integrating on both sides we get (1/2)log(2yc1)=log(z)+log(c2)    c2=yxzz2(1/2)\log(2y-c_1) =\log(z)+\log(\sqrt{c_2}) \implies c_2 = \frac{y-x-z}{z^2} .


Since x+y+z=c1x+y+z=c_1 and yxzz2=c2\frac{y-x-z}{z^2}=c_2 are independent solutions of the given differential equation, so a solution of the given differential equation is

c2=f(c1)    yxzz2=f(x+y+z)c_2= f(c_1) \implies \frac{y-x-z}{z^2} = f(x+y+z) ________(1)


On x2+y2=1,z=1x^2+y^2 = 1, z=1 , so put x=cos(t),y=sin(t)x= \cos{(t)}, y = \sin{(t)} and put in solution, we get

sin(t)cos(t)=f(cos(t)+sin(t))\sin{(t)}-\cos{(t)}=f(\cos{(t)}+\sin{(t)}) ________(2)

So, the solution of the given differential equation is given by (1), where the function is given by (2).


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