Answer to Question #287031 in Differential Equations for Ravi

Question

{F} Solve the first order linear inhomogeneous differential equation using the Bernoulli method

y^,-(2y/x)=1+(1/x)

Answer 1

Bernoulli equation is of the form $\displaystyle \frac{dy}{dx}+p(x)y=q(x)y^n$. The given DE has n=0 which reduces the standard Bernoulli equation to $\displaystyle \frac{dy}{dx}+p(x)y=q(x)$ which can be solved as a first order linear differential equation.

Now, the given DE is;

$\displaystyle y\prime-\left(\frac{2}{x}\right)y=1+\frac{1}{x}$

putting $y=uv$ into the given DE yields;

$\displaystyle uv\prime+vu\prime-\left(\frac{2}{x}\right)uv=1+\frac{1}{x}\\ \Rightarrow uv\prime+v\left[u\prime-\frac{2}{x}u\right]=1+\frac{1}{x}\cdots\cdots(1)\\$

Putting the coefficient of $v$ equal zero yields;

$\displaystyle u\prime-\frac{2}{x}u=0\Rightarrow\frac{du}{dx}=\frac{2u}{x}$

By method of separation of variables, we have;

$\displaystyle$$\displaystyle \frac{1}{u}\frac{du}{dx}=\frac{2}{x}\\ \Rightarrow\int\frac{1}{u}\frac{du}{dx}\ dx=\int\frac{2}{x}\ dx\Rightarrow \int\frac{1}{u}\ du=\int\frac{2}{x}\ dx\Rightarrow \ln u=\ln x^2+\ln c\\ \Rightarrow u=cx^2, \text{where c is an arbitrary constant.}$

Next substituting $u=cx^2$ into (1) yields;

$\displaystyle cx^2v\prime=1+\frac{1}{x}$

By method of separation of variables, we have;

$\displaystyle c\frac{dv}{dx}=\frac{1}{x^2}+\frac{1}{x^3}$

integrating both sides wrt $x$ yields;

$\displaystyle c\int\frac{dv}{dx}\ dx=\int\left(\frac{1}{x^2}+\frac{1}{x^3}\right)\ dx\Rightarrow c\int dv=\int\left(\frac{1}{x^2}+\frac{1}{x^3}\right)\ dx$

$\displaystyle cv=-\frac{1}{x}-\frac{1}{2x^2}+k\Rightarrow v=\frac{1}{c}\left(-\frac{1}{x}-\frac{1}{2x^2}+k\right)$ , where k is an arbitrary constant.

Thus,

$\displaystyle y=uv=cx^2\times \frac{1}{c}\left(-\frac{1}{x}-\frac{1}{2x^2}+k\right)=-x-\frac{1}{2}+kx^2=kx^2-x-\frac{1}{2}$