# Solving ODE Separable Equations with Cs

• tony873004
In summary, the C in the y part of the equation is not necessarily the same C as from the x part, and they ended up with only 1 C. They got their C out of the parenthesis to end up multiplying by the right side.
tony873004
Gold Member
I'm getting confused as to what to do with the C's.

## Homework Statement

$$\frac{{dy}}{{dx}} + 2xy = 0$$

## Homework Equations

Back of Book: $$y\left( x \right) = C\exp \left( { - x^2 } \right)$$

## The Attempt at a Solution

$$\begin{array}{l} \frac{{dy}}{{dx}} = - 2xy\,\,\,\, \Rightarrow \,\,\,\,\frac{{dy}}{y} = - 2x\,dx\,\,\,\, \Rightarrow \,\,\,\,\frac{1}{y}dy = - 2x\,dx \\ \\ \int_{}^{} {\frac{1}{y}dy} = \int_{}^{} { - 2x\,dx} \\ \\ \ln \left( y \right) = \frac{{ - 2x^2 }}{2} + C\,\,\,\, \Rightarrow \,\,\,\,\ln \left( y \right) = - x^2 + C\,\,\,\, \Rightarrow \,\,\,\, \\ \\ \exp \left( {\ln \left( {y + C} \right)} \right) = \exp \left( { - x^2 + C} \right)\,\,\,\, \Rightarrow \,\,\,\, \\ \\ y + C = \exp \left( { - x^2 + C} \right)\,\,\,\, \Rightarrow \,\,\,\,y = \exp \left( { - x^2 + C} \right) - C \\ \end{array}$$

The C from the y part is not necessarily the same C as from the x part, is it? How did they end up with only 1 C, and how did they get their C out of the parenthesis to end up multiplying by the right side?

When you integrate both sides, you get C1 on the y side & C2 on the x side. You can combine them, to make C2-C1=C leaving it on the x side.

Also, I don't see where you've typed out your last question. Kind of confusing me as well.

Maybe this will help. e^(x+y)=e^x e^y

tony873004 said:

## The Attempt at a Solution

$$\exp \left( {\ln \left( {y + C} \right)} \right)$$

Where did this $C$ come from?

If you want to include a constant of integration for $$\int\frac{dy}{y}$$ You should have $$\int\frac{dy}{y}=\ln(y)+C\neq\ln(y+C)$$

Also, you should label your constants so that you don't confuse them with each other:

$$\int\frac{dy}{y}=\ln(y)+C_1$$

$$\int -2x dx=-x^2+C_2$$

$$\implies \ln(y)+C_1=-x^2+C_2 \implies \ln(y)=-x^2 +(C_2-C_1)$$

but $(C_2-C_1)$ is just going to be some constant, you may as well call $C_3$

[tex]\implies \ln(y)=-x^2+C_3 \implies y=e^{-x^2+C_3}=e^{C_3}e^{-x^2}[/itex]

And then just call $C\equiv e^{C_3}$.

Thanks rocomath. I understand up to e^(-x^2) e^C . But how does that turn into Ce^(-x^2) ?
For example, 2^3 * 2^4 doesn’t turn into 4*2^3.
128 <> 32

e is a positive constant, C is a constant, positive constant^constant = constant. e^C = A - better?

Thanks, gabba^2hey! C3 did the trick!

## 1. What are separable equations in ODE?

Separable equations in ODE are equations of the form dy/dx = f(x)g(y), where the variables x and y can be separated and solved for independently. This type of equation can be solved by integrating both sides and using the initial conditions to find the constants of integration.

## 2. How do I know if an ODE is separable?

An ODE is separable if it can be written in the form dy/dx = f(x)g(y), where f(x) and g(y) are functions of x and y respectively. If the equation cannot be written in this form, then it is not separable and a different method must be used to solve it.

## 3. What is the general process for solving separable equations?

The general process for solving separable equations is as follows:

1. Separate the variables x and y on either side of the equation.
2. Integrate both sides of the equation with respect to x and y, respectively.
3. Add the constant of integration for each integration.
4. Use the initial conditions to solve for the constants of integration.

## 4. What is the role of the constant of integration in solving separable equations?

The constant of integration is a constant term that is added during the integration process. It is necessary because when taking the derivative of a constant, it becomes zero. The constant of integration allows us to account for all possible solutions to the separable equation.

## 5. Can I solve separable equations with initial conditions using the method of undetermined coefficients?

No, the method of undetermined coefficients cannot be used to solve separable equations with initial conditions. This method is only applicable to linear ODEs, while separable equations are nonlinear. The initial conditions must be used to solve for the constants of integration in the general solution of the separable equation.

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