# Examples On First Order Differential Equations

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Example -13

Solve the DE \begin{align}\frac{{dy}}{{dx}} + y\tan x = \cos x.\end{align}

Solution: Comparing this DE with the standard form of the linear DE \begin{align}\frac{{dy}}{{dx}} + Py = Q,\end{align} we see that

$P(x) = \tan x, \quad Q(x) = \cos x$

Thus, the I.F. is

\begin{align}I.F. = {e^{\int {\tan xdx} }} = {e^{\ln (\sec x)}} = \sec x\end{align}

Multiplying by $$\sec x$$ on both sides of the given DE, we obtain

$\sec x\frac{{dy}}{{dx}} + y\tan x\sec x = 1$

The left hand side is an exact differential:

\begin{align}&\qquad \;\frac{d}{{dx}}(y\sec x) = 1\\&\Rightarrow \quad d\left( {y\sec x} \right) = dx\end{align}

Integrating both sides, we obtain the solution to our DE as

$y\sec x = x + C$

Example - 14

Solve the DE\begin{align}\frac{{dx}}{{dy}} + \frac{x}{y} = {y^2}.\end{align}

Solution: The I.F. is

\begin{align}I.F. & = {e^{\int {P(y)dy} }} & ({\rm{y\,is\,the\,'independent'\,variable\,in\,this\,DE}})\\ & = {e^{\int {\frac{1}{y}dy} }}\\ & = {e^{\ln y}}\\ & = y\end{align}

Multiplying by the I.F. on both sides, we have

\begin{align} & \qquad \;\;y\frac{{dx}}{{dy}} + x = {y^3}\\&\Rightarrow \quad \frac{d}{{dy}}(xy) = {y^3}\\&\Rightarrow \quad d(xy) = {y^3}dy\end{align}

Integrating both sides gives

$xy = \frac{{{y^4}}}{4} + C$

Example – 15

Solve the DE  \begin{align}\frac{{dy}}{{dx}} = {x^3}{y^3} - xy.\end{align}

Solution: We have,

$\frac{{dy}}{{dx}} + xy = {x^3}{y^3}$

Note that since the RHS contains the term $${y^3},$$ this DE is not in the standard linear DE form. However, a little artifice can enable us to reduce this to the standard form.

Divide both sides of the equation by $${y^3}.$$

$\frac{1}{{{y^3}}}\frac{{dy}}{{dx}} + \frac{x}{{{y^2}}} = {x^3}\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots (1)$

Substitute      \begin{align}\frac{1}{{{y^2}}} = v\end{align}

\begin{align}& \Rightarrow \quad \frac{{ - 2}}{{{y^3}}}\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}\\&\Rightarrow \quad \frac{1}{{{y^3}}}\frac{{dy}}{{dx}} = \frac{{ - 1}}{2}\frac{{dv}}{{dx}} \qquad \qquad \ldots (2)\end{align}

Using (2) in (1), we have

\begin{align}&\qquad \;\frac{{ - 1}}{2}\frac{{dv}}{{dx}} + xv = {x^3}\\&\Rightarrow \quad \frac{{dv}}{{dx}} + ( - 2x)v = - 2{x^3} \qquad \qquad \cdots (3)\end{align}

This is now in the standard first-order linear DE form. The I.F. is

$I.F. = {e^{\int { - 2xdx} }} = {e^{ - {x^2}}}$

Thus, the solution to (3) is

\begin{align}& \qquad \;\;v \times I.F. = \int {Q(x) \times (I.F)dx} \\&\Rightarrow \quad v{e^{ - {x^2}}} = - 2\int {{x^3}{e^{ - {x^2}}}dx}\end{align}

Performing the integration on the RHS by the substitution $$t = - {x^2}$$ and then using integration by parts, we obtain

\begin{align}&\qquad \;\; v{e^{ - {x^2}}} = {e^{ - {x^2}}}({x^2} + 1) + C\\&\Rightarrow \quad \frac{1}{{{y^2}}}{e^{ - {x^2}}} = {e^{ - {x^2}}}({x^2} + 1) + C\end{align}

This is the required general solution to the DE.

This example also tells us how to solve a DE of the general form

$\frac{{dy}}{{dx}} + P(x)y = Q(x){y^n} \qquad \qquad \ldots (4)$

We divide by $${y^n}$$ on both sides:

$\frac{1}{{{y^n}}}\frac{{dy}}{{dx}} + P(x){y^{ - n + 1}} = Q(x)$

and then substitute $${y^{ - n + 1}} = v$$ and proceed as described in the solution above.

DEs that take the form in (4) are known as Bernoulli’s DEs.