Determine whether each first-order differntial equation is separable, linear, bo

pedzenekO 2021-09-20 Answered
Determine whether each first-order differntial equation is separable, linear, both or nether:
a) \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{e}^{{{x}}}{y}={x}^{{{2}}}{y}^{{{2}}}\)
b) \(\displaystyle{y}+{\sin{{x}}}={x}^{{{3}}}{y}'\)
c) \(\displaystyle{\ln{{x}}}-{x}^{{{2}}}{y}={x}{y}'\)
d) \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{\cos{{y}}}={\tan{{x}}}\)

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Expert Answer

Talisha
Answered 2021-09-21 Author has 3753 answers
Step 1
A first-order differential equation is said to be separable if it can be written as
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={f{{\left({x}\right)}}}{g{{\left({y}\right)}}},\ {f}\) and \(\displaystyle{g}\) are known functions.
A first-order differntial equation is said to be separable if it can be written as
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{p}{\left({x}\right)}{y}={q}{\left({x}\right)},\ {p}{\left({x}\right)}\) and \(\displaystyle{q}{\left({x}\right)}\) are continuouse functions.
a) Consider the differential equation \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{e}^{{{x}}}{y}={x}^{{{2}}}{y}^{{{2}}}\)
Rewrite the given differential equation as follows:
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}^{{{2}}}{y}^{{{2}}}-{e}^{{{x}}}{y}\)
\(\displaystyle={y}{\left({x}^{{{2}}}{y}-{e}^{{{x}}}\right)}\)
As the variables \(\displaystyle{x}\) and \(\displaystyle{y}\) cannot be separated and written as separate functions of \(\displaystyle{x}\) and \(\displaystyle{y}\), given equation is not separable.
Also, in the equation \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{e}^{{{x}}}{y}={x}^{{{2}}}{y}^{{{2}}},\ {p}{\left({x}\right)}={e}^{{{x}}}\) but right hand side is not a function of x alone. Hence, it is not a linear equation.
Therefore, the given differential equation is neither a separable equation nor a linear equation.
b) Consider the differential equation \(\displaystyle{y}+{\sin{{x}}}={x}^{{{3}}}{y}'\)
Rewrite the given differential equation as follows:
\(\displaystyle{y}+{\sin{{x}}}={x}^{{{3}}}{y}'\)
\(\displaystyle{y}+{\sin{{x}}}={x}^{{{3}}}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\)
\(\displaystyle{\frac{{{y}}}{{{x}^{{{3}}}}}}+{\frac{{{\sin{{x}}}}}{{{x}^{{{3}}}}}}={\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\)
\(\displaystyle{\frac{{{\sin{{x}}}}}{{{x}^{{{3}}}}}}-{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}-{\frac{{{y}}}{{{x}^{{{3}}}}}}\)
Thus, the given equation can be written as \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{\left({\frac{{-{1}}}{{{x}^{{{3}}}}}}\right)}{y}={\frac{{{\sin{{x}}}}}{{{x}^{{{3}}}}}}\) where
\(\displaystyle{p}{\left({x}\right)}={\frac{{-{1}}}{{{x}^{{{3}}}}}}\) and \(\displaystyle{q}{\left({x}\right)}={\frac{{{\sin{{x}}}}}{{{x}^{{{3}}}}}}\)
Hence, it is a linear equation.
Also, equation can be written as \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{1}}}{{{x}^{{{3}}}}}}{\left({y}+{\sin{{x}}}\right)}\)
As the variables \(\displaystyle{x}\) and \(\displaystyle{y}\) cannot be separated and written as separate functions of \(\displaystyle{x}\) and \(\displaystyle{y}\), given equation is not separable.
Therefore, the given differential equation is only linear equation.
Step 2
c) Consider the differential equation \(\displaystyle{\ln{{x}}}-{x}^{{{2}}}{y}={x}{y}'\)
Rewrite the given differential equation as follows:
\(\displaystyle{\ln{{x}}}-{x}^{{{2}}}{y}={x}{y}'\)
\(\displaystyle{\ln{{x}}}-{x}^{{{2}}}{y}={x}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\)
\(\displaystyle{\frac{{{\ln{{x}}}}}{{{x}}}}-{x}{y}={\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\)
\(\displaystyle{\frac{{{\ln{{x}}}}}{{{x}}}}={\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{x}{y}\)
Thus, the given equation can be written as \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{x}{y}-{\frac{{{\ln{{x}}}}}{{{x}}}}\) where \(\displaystyle{p}{\left({x}\right)}-{x}\) and \(\displaystyle{q}{\left({x}\right)}-{\frac{{{\ln{{x}}}}}{{{x}}}}\).
Hence, it is a linear equation.
Also, equation can be written as \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{1}}}{{{x}}}}{\left({\ln{{x}}}-{x}^{{{2}}}{y}\right)}.\)
As the variables \(\displaystyle{x}\) and \(\displaystyle{y}\) cannot be separated and written as separate functions of \(\displaystyle{x}\) and \(\displaystyle{y}\), given equation is not separable.
Therefore, the given differential equation is only linear equation.
d) Consider the differential equation \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{\cos{{y}}}={\tan{{x}}}\).
Here, \(\displaystyle{y}\) is the dependent variable and \(\displaystyle{x}\) is the independent variable.
Since, \(\displaystyle{\cos{{y}}}{1}-{\frac{{{y}^{{{2}}}}}{{{2}!}}}+{\frac{{{y}^{{{4}}}}}{{{4}!}}}+\cdots\), the dependent variable occurs in higher powers.
So, this is not a linear equation.
Also, the variables \(\displaystyle{x}\) and \(\displaystyle{y}\) cannot be separated and written as separate functions of \(\displaystyle{x}\) and \(\displaystyle{y}\), given equation is not separable.
The given differential equation cannot be written in any of the forms.
Therefore, the given differential equation is neither a separable equation nor a linear equation.
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