## DanielM_113 What is the general solution to the ODE /[dy/dx = 2y+1?/] Use separation of variables. Is y' + xy = x separable? 10 months ago 10 months ago

1. DanielM_113

$\int \frac{dy}{2y+1}=\int dx \\ \ 2 \ln y = x$

2. DanielM_113

What are the steps to the integration of the equation? $\int \frac{dy}{2y+1} \\ \frac{dt}{dy} = 2y+1 \\ \int dt = \int\\ \int \frac{dt}{t} = \ln t \\$

3. DanielM_113

Hmm, Im having trouble remembering this.

4. DanielM_113

$t=2y+1 \\ \frac{1}{t}=\frac{1}{2y+1} \\ \frac{dt}{dy}=2 \\$ Is this true? $\int\frac{dt}{t} = \int \frac{dy}{2y+1}\quad\mbox{?}$

5. DanielM_113

I need to visualize integration by substitution better.

6. DanielM_113

if I set $u=2y+1\\ \frac{\mathrm du}{\mathrm dy}= 2\\ \frac{1}{u}\frac{\mathrm du}{\mathrm dy}=\frac{1}{2y+1}2 \\ \frac{1}{2}\int \frac{\mathrm du}{u}=\int\frac{1}{2y+1}\mathrm dy \\ \int \frac{\mathrm du}{u} = \ln u = \ln (2y +1)\\ \int\frac{1}{2y+1}\mathrm dy =\frac{1}{2}\int \frac{\mathrm du}{u}=\frac{1}{2}\ln (2y +1)$

7. DanielM_113

$\int\frac{1}{2y+1}\mathrm dy = \int \mathrm dx \\ \frac{1}{2}\ln (2y +1) + c_1= x + c_2\\ \ln (2y +1) = 2x + 2c_2 -c_1\\ 2y+1=e^{2x+2c_2-c_1}\\ 2y+1=e^{2x+c_3}\\ y=\frac{1}{2}e^{c_3}e^{2x}\\ y=\frac{1}{2}Ce^{2x}$ This looks right for the first question.

8. DanielM_113

$\int\frac{1}{2y+1}\mathrm dy = \int \mathrm dx \\ \frac{1}{2}\ln (2y +1) + c_1= x + c_2\\ \ln (2y +1) = 2x + 2c_2 -c_1\\ 2y+1=e^{2x+2c_2-c_1}\\ 2y+1=e^{2x+c_3}\\ y=\frac{1}{2}(e^{c_3}e^{2x} - 1)\\ y=\frac{1}{2}(Ce^{2x} - 1)$ This looks right for the first question.

9. DanielM_113

I forgot the -1 on the previous one.

10. amistre64

$dy/dx = 2y+1$ is seperable in the sense that we can form $g(y)~dy = f(x)~dx$ $dy/dx = 2y+1$ $\frac{1}{2y+1}dy = dx$ $\int \frac{1}{2y+1}dy =\int dx$ $\frac12ln(2y+1) =x+C$ keep in mind the that +C is an all encompassing constant: therefore 2C and c1+c2 and the like are contained within it. $ln(2y+1) =2x+C$ $2y+1 =e^{2x+C}$ $2y =e^{2x+C}-1$ $y =\frac12e^{2x+C}-\frac12$ $y =\frac12Ce^{2x}-\frac12$ so tahts fine

11. amistre64

y' + xy = x dy/dx = x(1-y) dy/(1-y) = x dx that can be seperated as well

12. DanielM_113

The first one checks with my answer, thank you! And the second one looks great too.