General Solution of differential equation (x - y^2)dx +y(5x + y^2)dy

The general solution of the differential equation $\left( {x - {y^2}} \right)dx + y\left( {5x + {y^2}} \right)dy = 0$ is:

${\left( {{y^2} + x} \right)^4} = C\left| {{{\left( {{y^2} + 2x} \right)}^3}} \right|$
${\left( {{y^2} + 2x} \right)^4} = C\left| {{{\left( {{y^2} + x} \right)}^3}} \right|$
$\left| {{{\left( {{y^2} + x} \right)}^3}} \right| = C{\left( {2{y^2} + x} \right)^4}$
$\left| {{{\left( {{y^2} + 2x} \right)}^3}} \right| = C{\left( {2{y^2} + x} \right)^4}$

Solution:

Find the general solution of a given differential equation using the substitution method.

Given, $\left( {x - {y^2}} \right)dx + y\left( {5x + {y^2}} \right)dy = 0$

$\begin{gathered} \Rightarrow y\left( {5x + {y^2}} \right)dy = - \left( {\left( {x - {y^2}} \right)dx} \right) \hfill \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{ - \left( {x - {y^2}} \right)}}{{y\left( {5x + {y^2}} \right)}} \hfill \\ \end{gathered}$

$\Rightarrow y\frac{{dy}}{{dx}} = \frac{{\left( {{y^2} - x} \right)}}{{\left( {5x + {y^2}} \right)}}$ … (1)

Let ${y^2} = t$

On differentiation, we get

$\Rightarrow 2y\frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}$

Then, equation (1) becomes

$\Rightarrow \frac{1}{2}\frac{{dt}}{{dx}} = \frac{{\left( {t - x} \right)}}{{\left( {5x + t} \right)}}$ … (2)

Now, Substitute t = vx

On differentiation, we get

$\Rightarrow \frac{{dt}}{{dx}} = v + x\frac{{dv}}{{dx}}$

Then, equation (2) becomes

$\Rightarrow \frac{1}{2}\left\{ {v + x\frac{{dv}}{{dx}}} \right\} = \frac{{v - 1}}{{5 + v}}$

$\begin{gathered} \Rightarrow x\frac{{dv}}{{dx}} = \frac{{2v - 2}}{{5 + v}} - v \hfill \\ \Rightarrow x\frac{{dv}}{{dx}} = \frac{{ - 3v - {v^2} - 2}}{{5 + v}} \hfill \\ \Rightarrow x\frac{{dv}}{{dx}} = \frac{{ - (3v + {v^2} + 2)}}{{5 + v}} \hfill \\ \end{gathered}$

$\Rightarrow \frac{{5 + v}}{{3v + {v^2} + 2}}dv = - \frac{{dx}}{x}$ … (3)

Now, integrate equation (3)

$\Rightarrow \int {\frac{{5 + v}}{{3v + {v^2} + 2}}} dv = \int { - \frac{{dx}}{x}}$

Integrate using partial faction

On Partial Fraction, $\frac{{5 + v}}{{3v + {v^2} + 2}} = \frac{4}{{v + 1}} - \frac{3}{{v + 2}}$

$\begin{gathered} \Rightarrow \int {\left( {\frac{4}{{v + 1}} - \frac{3}{{v + 2}}} \right)dv = \int { - \frac{{dx}}{x}} } \hfill \\ \Rightarrow \int {\frac{4}{{v + 1}}} dv - \int {\frac{3}{{v + 2}}} dv = - \int {\frac{{dx}}{x}} \hfill \\ \Rightarrow 4\ln \left| {v + 1} \right| - 3\ln \left| {v + 2} \right| = - \ln x + \ln c \hfill \\ \end{gathered}$

$\Rightarrow \left| {\frac{{{{\left( {v + 1} \right)}^4}}}{{{{\left( {v + 2} \right)}^3}}}} \right| = \frac{c}{x}$ … (4)

Now, $t = vx \Rightarrow v = \frac{t}{x} = \frac{{{y^2}}}{x}$

Then equation (4) becomes

$\Rightarrow \left| {\frac{{{{\left( {\frac{{{y^2}}}{x} + 1} \right)}^4}}}{{{{\left( {\frac{{{y^2}}}{x} + 2} \right)}^3}}}} \right| = \frac{c}{x}$

$\Rightarrow {\left( {{y^2} + x} \right)^4} = C\left| {{{\left( {{y^2} + 2x} \right)}^3}} \right|$

Final Answer:

Hence, Option (A) is correct.

Solution:

Final Answer:

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