Solving ODE Separable Equations with Cs

[tony873004]

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

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

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?

Thanks in advance!

 

[rocomath]

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.

 

[rocomath]

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

 

[gabbagabbahey]

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]<br /> <br /> And then just call $C\equiv e^{C_3}$.[/tex]

 

[tony873004]

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

 

[NoMoreExams]

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

 

[tony873004]

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