Differential Equation ODE Solution help.

[Physics345]

Homework Statement

y(x+y+1)dx+(x+2y)dy=0

Relevant Equations

(My-Nx)/n

dM/dY = x+2y+1 dN/dx = 1

(My-Nx)/n = 1 Integrating Factor => e^∫1dx= e^x

(xye^x+ye^x+ye^x)dx + (xe^x+2ye^x)dy = 0

dM/dY =xye^x+e^x+2ye^x dN/dx = xye^x+e^x+2ye^x Exact

∫dF/dy * dy = ∫ (xe^x+2ye^x)dy

F = xy*e^x + y^2*e^x + c(x)

dF/dx = xy*e^x + y*e^x + y^2 * e^x + c'(x)

c'(x) = 0

c(x) = c

Therefore, the general solution to the ODE is xye^x + y^2 * e^x + c = 0

Did I miss something here? The doubt stems from c'(x) = 0

Is there any way I can confirm the answers to my ODE solutions?

Thanks for the help everyone.

 

[Mark44]

Problem Statement: y(x+y+1)dx+(x+2y)dy=0
Relevant Equations: (My-Nx)/n

dM/dY = x+2y+1 dN/dx = 1

(My-Nx)/n = 1 Integrating Factor => e^∫1dx= e^x

(xye^x+ye^x+ye^x)dx + (xe^x+2ye^x)dy = 0

dM/dY =xye^x+e^x+2ye^x dN/dx = xye^x+e^x+2ye^x Exact

∫dF/dy * dy = ∫ (xe^x+2ye^x)dy

F = xy*e^x + y^2*e^x + c(x)

dF/dx = xy*e^x + y*e^x + y^2 * e^x + c'(x)

c'(x) = 0

c(x) = c

Therefore, the general solution to the ODE is xye^x + y^2 * e^x + c = 0

Did I miss something here? The doubt stems from c'(x) = 0

Is there any way I can confirm the answers to my ODE solutions?

Thanks for the help everyone.

Differentiate your solution with respect to x, using implicit differentiation. Doing this, you're assuming that y is implicitly a function of x alone. When I did that, I was able to get back to your original differential equation.

 

[Physics345]

Differentiate your solution with respect to x, using implicit differentiation. Doing this, you're assuming that y is implicitly a function of x alone. When I did that, I was able to get back to your original differential equation.

xy*e^x + ye^x + xe^x * dy/dx + y^2 * e^x + 2ye^x * dy/dx = 0 divide by e^x

xy + y + x * dy/dx + y^2 + 2y * dy/dx = 0

y(y+x+1)dx + (x+2y)dy = 0

Genius! You're a lifesaver, thanks.

 

[epenguin]

Even people who are not geniuses can do this and always should!