Answer to Question #153486 in Differential Equations for Ashweta Padhan

Question

Find the integral surface passing through the initial data curve C: x₀= -1, y₀= s, z₀= (s)^1/2 of the equation (x+2)p +2yq =2z

Answer 1

we have to find the integral surface of the equation :

$(x+2)p+2yq=2z..........(1)$

From observation we can see that it is "Lagrange's partial differential equation" of type

$Pp+Qq=R$

So, from equation$(1)$ we get Lagrange's auxiliary equation

$\frac{dx}{x+2}=\frac {dy}{2y}=\frac{dz}{2z}\,.............(2)$

Now we take

$\frac{dx}{x+2}=\frac{dy}{2y}$

$\implies\intop\frac{dx}{x+2}=\intop\frac{dy}{2y}\\\implies\ln(x+2)=\frac{1}{2}\ln y+\ln c_1\\\implies\ln(x+2)-\ln(\sqrt{y})=ln c_1\\\implies\frac{x+2}{\sqrt{y}}=c_1............(3)$

we will take another equation

$\frac{dy}{2y}=\frac{dz}{2z}\\\implies\intop\frac{dz}{2z}=\intop\frac{dy}{2y}\\\implies\frac{1}{2}\ln z-\frac{1}{2}ln y=\ln\sqrt{c_2}\\\implies\sqrt{\frac{z}{y}}=\sqrt{c_2}\\\implies\frac{z}{y}=c_2..............(4)$

Now it is given that

$x_0=-1;y_0=s_0;z_0=\sqrt{s_0}\\$

putting the values of $x_0;y_0;z_0$ in the equation (3) &(4) we get

$c_1=\sqrt{\frac{1}{s_0}}$ & $c_2=$ $\sqrt{\frac{1}{s_0}}$

now putting the values of C₁& C₂ we get

$\frac{x+2}{\sqrt{y}}=\sqrt{\frac{1}{s_0}}.......(5)$ & $\frac{z}{y}=\sqrt{\frac{1}{s_0}}..........(6)$

now subtracting equation (5) to equation (6) we will get

$\frac{x+2}{\sqrt{y}}=$ $\frac{z}{y}$

Hence

$\frac{x+2}{\sqrt{y}}=$ $\frac{z}{y}$ ; is the equation of the integral surface of the given equation.