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$$y' = \frac{-y-x}{x}, y' = \frac{-y}{x} -1$$ $$F(v) = -v -1$$ Since $$y = xv$$ then $$y' = v + xv'$$ Therefore: $$v+xv' = -v-1$$ $$xv' = -2v-1$$ $$\frac{-dv}{2v+1} = \frac{dx}{x}$$ $$-\frac{1}{2}ln\bigg |1+2v\bigg|=ln|x| + C$$ $$ln|1+2v| = -2ln|x| -2C$$ $$1+2v = \pm (x^{-2}*e^{-2C})$$ So here is where I get anxious, I'm not sure what to do with with the constants and the plus/minus from the LHS absolute value sign. But here's what I did and what we learned in class:

$$1+2v = \frac{D}{x^{2}}$$ where $D = \pm e^{-2C}$ and so $v =\frac{D}{2x^{2}} - \frac{1}{2}$

Do I substitute $v = \frac{y}{x}$ back in? So $ \frac{y}{x}= \frac{D}{2x^{2}} - \frac{1}{2}$

$$y = \frac{xD}{2x^{2}} - \frac{x}{2}$$ $$y = \frac{D}{2x} - \frac{x}{2}$$

Plugging in the initial condition where y(1) = 1:

$$1 = \frac{D}{2} - \frac{1}{2}$$ $$D = 3$$

Therefore, the general solution is...?

$$y(x) = \frac{3}{2x} - \frac{x}{2}$$

The thing I'm confused about is, are we solving for $D$ since we substituted $D$ for $e^{-2C}$?

Or does the constant variable doesn't matter?

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    $\begingroup$ yor result is right i got the same equation $\endgroup$ – Dr. Sonnhard Graubner Feb 4 '18 at 5:02
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    $\begingroup$ I notice you have no answers accepted yet. In case you find that the answer was helpful and complete you can accept it by clicking the checkmark on left side. Each accepted answer will give you +2 reputation points (except for the case of you answering your own question) $\endgroup$ – Yuriy S Feb 4 '18 at 11:05

Your result is correct. Your final result does not represent the general solution anymore, since you have included the initial conditions, so the solution you got is a specific solution to this differential equation, with the initial conditions you have specified. There is no more constant to solve for, since you have done that already.


$$y'x+y+x=0$$ $$(xy)'+x=0$$ Simply integrate $$xy =-\int xdx=-\frac {x^2} 2 +K$$ $$y =-\frac {x} 2 +\frac Kx$$ then since $y(1)=1 \to K= \frac 32$ $$y =-\frac {x} 2 +\frac {3}{2x}$$ So your answer is correct..


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