Question

In problems 23–28 Find an explicit solution to the given initial-value problem.

${\text{x}}^{\text{2}}\frac{\text{dy}}{\text{dx}}\text{= y - xy , y ( - 1) = - 1}$

Short answer

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

Step-by-step solution

Definition of "separable equation"

A first-order differential equation of the form $\frac{dy}{dx}=g\left(x\right)h\left(y\right)$is said to be separable or to have separable variables.

Separate the variables

The original Initial Value Problem

$x^2\frac{dy}{dx}=y-xy,y(-1)=-1$

Factor out

$x^2\frac{dy}{dx}=y(1-x)$

Separate your variables

$\frac{dy}{y}=\frac{1-x}{x^2}dx$

Integration

Integrate.

$$\begin{gathered} \int \frac{dy}{y}=\int \frac{1-x}{x^2}dx \\ \int \frac{dy}{y}=\int \frac{1}{x^2}dx-\int \frac{1}{x}dx \\ \ln \left|y\right|=-\frac{1}{x}-\ln \left|x\right|+C \end{gathered}$$

Put the natural logs on the same side.

$\ln \left|y\right|+\ln \left|x\right|=C-\frac{1}{x}$

By log rules, the log of a product is the sum of two logs. Thus, we can combine the two.

$\ln \left|yx\right|=C-\frac{1}{x}$

Rise to the power of each side to eliminate the log.

$e^{\ln \left|yx\right|}=e^{C-x^{-1}}$

The absolute values can be removed if the other side is made $\pm$Since $e^{\text{C}}$will be constant, it can be replaced with the constant C where C ranges over all real numbers.

$\left|yx\right|=e^C{e}^{-x^{-1}}$

$yx=C{e}^{-x^{-1}}$

Remember, we have the parameter$y\left(-1\right)=\left(-1\right)$

This will allow us to solve for C

$y=\frac{C{e}^{-x^{-1}}}{x}$

Simplification

Simplify.

or$$\begin{gathered} -1=\frac{C{e}^{-{(-1)}^{-1}}}{-1} \\ 1=Ce \end{gathered}$$

We have solved for our constant C

$\frac{1}{e}=C$or$C=e^{-1}$

Substitute$e^{-1}$ in for C in our equation, which was already solved for and simplified.

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

By exponent rules, if you have two exponents of the same base, the product is the sum of the exponents.

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

Hence, the solution is$\text{y =}\frac{{\text{e}}^{\text{-}\left(\text{1 +}\frac{\text{1}}{\text{x}}\right)}}{\text{x}}$