Answer to Question #310692 in Differential Equations for Ronan

Question

find the complete integral and singular integral of 4(1+x³)=9z⁴pq.

Answer 1

4(1+x³)=9z²pq

f(x,y,z,p, q)=4x³- 9z²pq+4=0.............(1)

$\frac{dp}{\frac{dF}{dx}+P\frac{dF}{dz}}=\frac{dq}{\frac{dF}{dy}+q\frac{dF}{dz}}=\frac{dz}{-P\frac{dF}{dp}-q\frac{dF}{dq}} =\frac{dx}{-\frac{dF}{dx}} =\frac{dy}{-\frac{dF}{dq}}$

$\frac{dp}{12x^2 -18zp^2q}=\frac{dq}{-18zpq^2}=\frac{dz}{9z^2pq-9z^2pq} =\frac{dx}{-9z^2q}=\frac{dy}{-9z^2p}$

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Solving this we get dz=0

Integrating (1), we equate q=C₁

$4x^3-9z^2pC1+4=0$

$P=\frac{4x^3+4}{9z^2C1}$

$dz= pdx+qdy$

$dz= \frac{4(x^3+4)}{9z^2C1}dx+C1dy$

$dz-C1dy= \frac{4(x^3+1)}{9z^2C1}dx$

$d(z-C1y)=\frac{4(x^3+1)}{9z^2C1}$

Integrating this we get,

$log(z-C1y)=log9z^2C1+log C2$

C_{2 is an arbitrary constant}

$Z-C1y =\frac{4}{9z^2C1}+C2.............(2)$

(2) is our complete integral

To get the singular integration we add 1 and 2

Differentiate 2 with respect to c₁ and c₂ respectively we get,

$y=9z^2C2$

$0=9z^2c1, c1=0 C2=\frac{y}{9z^2}$

z=0

Therefore y=c2