Question #294784

Solve (y ^ 2 * z)/x * p + xzq = y ^ 2 .


1
Expert's answer
2022-02-08T10:59:01-0500

Given (y2+z2)pxyq+xz=0(y^2+z^2)p - xyq + xz = 0

    (y2+z2)pxyq=xz\implies (y^2+z^2)p - xyq =- xz

So, by Lagrange's method, we have

dxy2+z2=dyxy=dzxz\frac{dx}{y^2+z^2} = \frac{dy}{-xy} = \frac{dz}{-xz} ______________(1)

    xdx+ydy+zdzxy2+xz2xy2xz2=dzxz\implies \frac{x dx + ydy + zdz}{xy^2 + xz^2 - xy^2 - xz^2} = \frac{dz}{-xz}

    xdx+ydy+zdz=0\implies x dx + ydy + zdz=0

So by integration, we get x2+y2+z2=c1x^2 + y^2 + z^2 = c_1 _____________(2)

Also from last two parts of equation (1), we have

dyxy=dzxz\frac{dy}{-xy} = \frac{dz}{-xz}

    dyy=dzz\implies \frac{dy}{y} = \frac{dz}{z}

So by integration, we get

ln(y)=ln(z)+ln(c2)\ln(y) = \ln(z) + \ln(c_2)

    yz=c2\implies \frac{y}{z} = c_2 __________(3)

Now, from (2) and (3), solution of given differential equation is

c2=f(c1)    yz=f(x2+y2+z2)c_2 = f(c_1) \\ \implies \frac{y}{z} = f(x^2 + y^2 + z^2) is the solution.


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