Question

Find if the following first order differential equations seperable,linear,exact,almost exact,homogeneous.(x)(e^y)(dy/dx) = e^(-2x) + e^(y-2x)

Differential equations
ANSWERED
asked 2021-09-15
Find if the following first order differential equations seperable, linear, exact, almost exact, homogeneous, or Bernoulli. Rewrite the equation into standard form for the classification it fits.
\(\displaystyle{\left({x}\right)}{\left({e}^{{y}}\right)}{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}={e}^{{-{2}{x}}}+{e}^{{{y}-{2}{x}}}\)

Expert Answers (1)

2021-09-16
Separation of variables: If in an equation, it is possible to get all the functions of x and dx to one side and all the functions of y and dy to the other, the variables are said to be separable.
Linear Differential Equation: A differential equation is called linear if every dependent variable and every derivative involved occurs in the first degree only, and no products of dependent variables and/or derivatives occur.
Exact Differential Equation: The necessary and efficient for the differential equation \(\displaystyle{\left({M}\right)}{\left.{d}{x}\right.}+{\left({N}\right)}{\left.{d}{y}\right.}={0}\) to be exact is
\(\displaystyle{\frac{{\partial{M}}}{{\partial{y}}}}={\frac{{\partial{n}}}{{\partial{x}}}}\)
Homogeneous equation: A differential equation of first order and first N degree is said to be homogeneous if it can be put in the form
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={f{{\left({\frac{{{y}}}{{{x}}}}\right)}}}\)
Or, equations of the type \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{A}{\left({x},{y}\right)}}}{{{B}{\left({x},{y}\right)}}}}\) where
\(\displaystyle{A}{\left(\lambda{x},\lambda{y}\right)}=\lambda^{{d}}{A}{\left({x},{y}\right)}\)
\(\displaystyle{B}{\left(\lambda{x},\lambda{y}\right)}=\lambda^{{d}}{B}{\left({x},{y}\right)}\)
are homogeneous equations of degree d.
Bernoulli’s Equation: An equation of the form
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{P}{y}={Q}{y}^{{n}}\)
where P and Q are constants or functions of x alone (amd not of y) and n is constant except 0 and 1, is called a Bernoulli’s differential equation.
The given differential equation is \(\displaystyle{\left({x}\right)}{\left({e}^{{y}}\right)}{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}={e}^{{-{2}{x}}}+{e}^{{{y}-{2}{x}}}\)
It can be written as \(\displaystyle{\left({x}\right)}{\left({e}^{{y}}\right)}{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}={e}^{{-{2}{x}}}+{e}^{{{y}-{2}{x}}}\)
implies \(\displaystyle{\left({x}\right)}{\left({e}^{{y}}\right)}{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}={e}^{{-{2}{x}}}-{e}^{{y}}{e}^{{-{2}{x}}}\)
implies \(\displaystyle{\left({x}\right)}{\left({e}^{{y}}\right)}{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}={e}^{{-{2}{x}}}{\left({1}-{e}^{{y}}\right)}\)
implies \(\displaystyle{\frac{{{e}^{{y}}}}{{{1}-{e}^{{y}}}}}{\left.{d}{y}\right.}=\frac{{1}}{{x}}{e}^{{-{2}{x}}}{\left.{d}{x}\right.}\)
The function of x, dx and y,dy can be written on different sides, so the equation is separable
It has products of dependent variables, so this is not a linear differential equation.
Here \(\displaystyle{M}=\frac{{1}}{{x}}{e}^{{-{2}{x}}}{\quad\text{and}\quad}{N}={\frac{{{e}^{{y}}}}{{{1}-{e}^{{y}}}}}\)
Now, \(\displaystyle{\frac{{\partial{M}}}{{\partial{y}}}}={0}{\quad\text{and}\quad}{\frac{{\partial{n}}}{{\partial{x}}}}={0}\)
So, \(\displaystyle{\frac{{\partial{M}}}{{\partial{y}}}}={\frac{{\partial{n}}}{{\partial{x}}}}\). This is an exact equation
The given equation can not be written in the form \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={f{{\left({\frac{{{y}}}{{{x}}}}\right)}}}.\)
So, it is not a homogeneous equation
\(\displaystyle{\left({x}\right)}{\left({e}^{{y}}\right)}{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}={e}^{{-{2}{x}}}+{e}^{{{y}-{2}{x}}}\)
implies \(\displaystyle{\left({x}\right)}{\left({e}^{{y}}\right)}{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}={e}^{{-{2}{x}}}+{e}^{{y}}{e}^{{{2}{x}}}\)
implies \(\displaystyle{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}-{\frac{{{1}}}{{{x}}}}{e}^{{-{y}}}{e}^{{-{2}{x}}}=-\frac{{1}}{{x}}{e}^{{-{2}{x}}}\)
which is not the required form. So this is not a Bernoulli's equation.
16
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours
...