Question

singular and general solution 

e^3x(p-1)+p^3e^2y=0;X=e^x ,Y=e^y by substitution

Answer 1

$$e^{3x}(p-1)+p^3e^{2y}=0$$

$$e^y(1-p)=p^3e^{3(y-x)}$$

Use the substitution $X=e^x, Y=e^y$

$$dX=e^xdx, dY=e^y dy$$

$$\dfrac{dy}{dx}=\dfrac{e^x}{e^y}\cdot\dfrac{dY}{dX}=>p=\dfrac{X}{Y}\cdot\dfrac{dY}{dX}$$

$$Y(1-\dfrac{X}{Y}\cdot\dfrac{dY}{dX})=(\dfrac{X}{Y}\cdot\dfrac{dY}{dX}\cdot\dfrac{Y}{X})^3$$

$$Y-X\dfrac{dY}{dX}=(\dfrac{dY}{dX})^3$$

The Clairaut equation 

$$Y=XP+P^3, P=\dfrac{dY}{dX}$$

Differentiating in $X,$ we have

$$Y=XP+P^3$$

$$dY=XdP+PdX+3P^2dP$$

Replace $dY$ with $PdX$ to obtain:

$$PdX=XdP+PdX+3P^2dP$$

$$dP(X+3P^2)=0$$

By equating the first factor to zero, we have

$$dP=0=>P=C$$

Now we substitute this into the differential equation.

The general solution is given by

$$Y=CX+C^3, C\text{ is arbitrary constant}$$

By equating the second term to zero we find that

$$X+3P^2=0=>X=-3P^2$$

This gives us the singular solution of the differential equation in parametric form:

$$X=-3P^2$$

$$Y=XP+P^3$$

By eliminating $P$ from this system, we get the equation of the integral curve:

$$P=\pm\sqrt{\dfrac{-X}{3}}, Y=\pm\dfrac{2}{3}X\sqrt{\dfrac{-X}{3}}$$

Since $X=e^x, Y=e^y,$ then the singular solution of the differential equation does not exist.

The general solution is given by

$$e^y=Ce^x+C^3, C\text{ is arbitrary constant}$$