Question

Plz use Differential equation method to solve this:

Answer 1

Answer:

Step-by-step explanation:

We have the differential equation $y' = y + \frac{x}{y}$ with initial conditions $y(0)=1$.

First, notice that the equation can be rewritten as

$y'-y =xy^{-1}$,

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

$y'+p(x)y=q(x)y^n$.

In this particular case we have $n=-1$. This kind of equation is solved by the change of variable $z=y^{1-n}$. In our exercise we get $z=y^{1-(-1)}=y^2$. Now we take derivatives and get

$z'=2yy'$ which es equivalent to $\frac{z'}{2y}=y'$.

Then, we substitute the value of $y'$ we have obtained in the original equation:

$\frac{z'}{2y}-y = xy^{-1}$.

The next step is to multiply the whole equation by $2y$, in order to eliminate the denominator of $z'$. Thus,

$z' -2y^2=2x$.

Recall that $y^2=z$, then

$z' -2z=2x$.

This last equation is a linear equation, which has general solution

$z(x) = \exp\left(-\int(-2)dx\right)\left(\int 2x \exp\left(\int(-2)dx + C\right)\right)$.

So, let us calculate the integral that appear in the formula:

$\int(-2)dx = -2x$

$\int 2x e^{-2x}dx = -\frac{\left(2x+1\right)e^{-2x}}{2}$.

Then, the solution for $z$ is

$z(x) = e^{2x}\left(-\frac{\left(2x+1\right)e^{-2x}}{2} + C\right) = -\frac{\left(2x+1\right)}{2} + Ce^{2x}$.

Now, we return the change of variable:

$(y(x))^2 =-\frac{\left(2x+1\right)}{2} + Ce^{2x}$.

The last step is to find the value of the constant $C$. In order to do this, substitute the initial value:

$(y(0))^2 = 1 =\frac{\left(2\cdot 0+1\right)}{2} + Ce^{2\cdot 0} = -\frac{1}{2} + C$.

Thus, we have the equation

$1=-\frac{1}{2} + C$ that gives $C=\frac{3}{2}$.

Therefore,

$(y(x))^2 = -\frac{\left(2x+1\right)}{2} + \frac{3}{2}e^{2x}$.