Find an explicit solution of the given initial-value problem

[fd25t6]

Hello all, hoping to get a bit of help with a diff eq problem that goes like this:

Find an explicit solution of the given initial-value problem,

x^2*(dy/dx)=y-xy ; y(-1) = -6

I separated my variables to get:

S (1-x)/x^2 dx = S 1/y dy

Integration of left via partial fractions:

S 1/x^2 dx - S 1/x dx = -1/x - ln|x|

Integration of right :

S 1/y dy = ln|y|

therefore :

-1/x - ln|x| = ln|y| + C

Now this is the first Calc class I have taken in a while so I am assuming more than one error here:

e^(-1/x) -x = y + e^C
y = e^(-1/x) -e^C - x
y = e^(1/x - C) -x

Did I solve for y correctly here? If not is my issue in the integration? Also at what point is it best to solve for C in this equation?

Thank you in advance for any assistance.

 

[Char. Limit]

Hello all, hoping to get a bit of help with a diff eq problem that goes like this:

Find an explicit solution of the given initial-value problem,

x^2*(dy/dx)=y-xy ; y(-1) = -6

I separated my variables to get:

S (1-x)/x^2 dx = S 1/y dy

Integration of left via partial fractions:

S 1/x^2 dx - S 1/x dx = -1/x - ln|x|

Integration of right :

S 1/y dy = ln|y|

therefore :

-1/x - ln|x| = ln|y| + C

You had it right up to here. That's definitely good.

Now this is the first Calc class I have taken in a while so I am assuming more than one error here:

e^(-1/x) -x = y + e^C
y = e^(-1/x) -e^C - x
y = e^(1/x - C) -x

Did I solve for y correctly here? If not is my issue in the integration? Also at what point is it best to solve for C in this equation?

Thank you in advance for any assistance.

You're kinda applying the exponential operator all wrong here. Your first step should look like

e^{-\frac{1}{x} - ln|x|} = e^{ln|y| + C}

From there, you'll want to separate the terms on the left via the exponential rule:

e^{a+b} = e^a e^b

You should get something akin to e^{-1/x} \times \frac{1}{x} on the left side, and of course e^C y on the right side. From there, it should be rather easy to simplify and plug in your initial condition to solve for e^C.