Answer:
Step-by-step explanation:
We have the differential equation [tex] y' = y + \frac{x}{y}[/tex] with initial conditions [tex]y(0)=1[/tex].
First, notice that the equation can be rewritten as
[tex] y'-y =xy^{-1}[/tex],
which is a Bernoulli equation. Once we have recognized the type of the equation we know how to continue. Recall that a Bernoulli equation has the general form
[tex]y'+p(x)y=q(x)y^n[/tex].
In this particular case we have [tex]n=-1[/tex]. This kind of equation is solved by the change of variable [tex]z=y^{1-n}[/tex]. In our exercise we get [tex]z=y^{1-(-1)}=y^2[/tex]. Now we take derivatives and get
[tex]z'=2yy'[/tex] which es equivalent to [tex]\frac{z'}{2y}=y'[/tex].
Then, we substitute the value of [tex]y'[/tex] we have obtained in the original equation:
[tex]\frac{z'}{2y}-y = xy^{-1}[/tex].
The next step is to multiply the whole equation by [tex]2y[/tex], in order to eliminate the denominator of [tex]z'[/tex]. Thus,
[tex]z' -2y^2=2x[/tex].
Recall that [tex]y^2=z[/tex], then
[tex]z' -2z=2x[/tex].
This last equation is a linear equation, which has general solution
[tex] z(x) = \exp\left(-\int(-2)dx\right)\left(\int 2x \exp\left(\int(-2)dx + C\right)\right)[/tex].
So, let us calculate the integral that appear in the formula:
[tex]\int(-2)dx = -2x[/tex]
[tex]\int 2x e^{-2x}dx = -\frac{\left(2x+1\right)e^{-2x}}{2}[/tex].
Then, the solution for [tex]z[/tex] is
[tex]z(x) = e^{2x}\left(-\frac{\left(2x+1\right)e^{-2x}}{2} + C\right) = -\frac{\left(2x+1\right)}{2} + Ce^{2x}[/tex].
Now, we return the change of variable:
[tex](y(x))^2 =-\frac{\left(2x+1\right)}{2} + Ce^{2x}[/tex].
The last step is to find the value of the constant [tex]C[/tex]. In order to do this, substitute the initial value:
[tex] (y(0))^2 = 1 =\frac{\left(2\cdot 0+1\right)}{2} + Ce^{2\cdot 0} = -\frac{1}{2} + C[/tex].
Thus, we have the equation
[tex]1=-\frac{1}{2} + C[/tex] that gives [tex]C=\frac{3}{2}[/tex].
Therefore,
[tex](y(x))^2 = -\frac{\left(2x+1\right)}{2} + \frac{3}{2}e^{2x}[/tex].
