# Homework Help: Dif. eq. problem - dont know why it is wrong

1. Feb 4, 2006

### UrbanXrisis

dif. eq. problem -- dont know why it is wrong

Solve for y: $$x \frac{dy}{dx} + 6y=5x$$

with y(1)=2

$$\frac{dy}{dx} =5 - 6\frac{y}{x}$$
$$\frac{6y}{x} \frac{dy}{dx} =5$$
$$3y^2=\frac{5}{2}x^2 +C$$
$$y= \sqrt{\frac{5x^2}{6}+\frac{C}{3}}$$

solve for when y(1)=2

$$4=\frac{5}{6}+\frac{C}{3}$$
$$C=\frac{19}{2}$$

so..

$$y= \sqrt{\frac{5x^2}{6}+\frac{19}{6}}$$

what did i do wrong? because this is not the answer

Last edited: Feb 4, 2006
2. Feb 4, 2006

### d_leet

y/x does not equal (dy)/(dx), You need to divide by x and then find an integrating factor.

3. Feb 4, 2006

### UrbanXrisis

sorry, that was typo! could you check it now?

4. Feb 4, 2006

### d_leet

It looks even worse, you somehow make addition into multiplaication. I already told you what you need to do. You need to divide everything by x to get the y' by itself and then find an integrating factor, you should know how to do that if you're being given this kind of problem because it certainly isn't separable,

5. Feb 4, 2006

### UrbanXrisis

oh thanks. by the way, when I am given a dif. eq. and they ask me "What are the constant solutions of this equation? "

what exactly are do they want me to find?

6. Feb 4, 2006

### ek

As said before, you're taking the wrong approach.

You need to divide it by x to get it into standard form or whatever it's called, and then get an integrating factor.

I get an answer of y = 5x/7 +c/x6

Last edited: Feb 4, 2006
7. Feb 4, 2006

### d_leet

Umm... I think they might mean for you to find the solutions that are just a constant function ie. y=c c is just some constant.

8. Feb 4, 2006

### d_leet

You forgot to divide c by x6

9. Feb 4, 2006

### ek

Ya. Actually I forgot to put in period and added it in haphazardly without thinking. I'm quite absent minded some times. I'll edit my post. I edited my last post.

Last edited: Feb 4, 2006
10. Feb 4, 2006

### Valhalla

hmmm

$$x\frac{dy}{dx}+6y=5x$$

$$\frac{dy}{dx}+\frac{6}{x}y=\frac{5}$$

$$e^{6\int\frac{dx}{x}}=x^6$$

multiply through by integrating factor

$$x^6\frac{dy}{dx}+6x^5y=5x^6$$

then integrate both sides

$$x^6y=\frac{5x^7}{7}+C$$

Divide through by x^6

$$y=\frac{5x}{7}+Cx^{-6}$$

do you see where your mistake is?

now solve for C

$$y(1)=2$$

$$2=\frac{5}{7}1+C$$

$$2-\frac{5}{7}=C$$

$$\frac{9}{7}=C$$

so the specific solution is

$$y=\frac{5x}{7}+\frac{9}{7}x^{-6}$$

Last edited: Feb 4, 2006
11. Feb 4, 2006

### d_leet

The term on the right hand side should be 5x not 5 so your answer is wrong as well.

12. Feb 4, 2006