 # 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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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.