Question

Find the general solution for the following differential equation: $\left( {x + 2} \right)\dfrac{{dy}}{{dx}} = {x^2} + 5x - 3$ where, $x \ne 2$.
(a) None of these
(b) $y = \dfrac{{{x^{_2}}}}{2} + 3x + 9\log \left( {x + 2} \right) + C$
(c) $y = \dfrac{{{x^{_2}}}}{2} + 3x - 9\log \left( {x + 2} \right) + C$
(d) $y = \dfrac{{{x^{_2}}}}{2} - 3x - 9\log \left( {x + 2} \right) + C$

Answer 1

Hint : The given problem revolves around the concept's differential equation (collaborated with both derivatives as well as integration terms). First of all, solving a given equation by taking ‘y’ terms and ‘x’ terms on one side i.e. on L.H.S. and R.H.S. respectively. Then, taking the integration, solving the equation in accordance with the rules/formulae of integration (by dividing the respective quadratic equation with the other linear equation in the solution), to obtain the desired solution.

Complete step-by-step answer :
Since, we have given the equation that
$\left( {x + 2} \right)\dfrac{{dy}}{{dx}} = {x^2} + 5x - 3$
Where, $x \ne 2$
Simplifying the given equation that is by dividing by $x + 2$,we get
$\dfrac{{dy}}{{dx}} = \dfrac{{{x^2} + 5x - 3}}{{x + 2}}$
Now, taking the derivative term that is with respect to ‘$x$’ to the above equation that is ‘$dx$’, we get
$dy = \left( {\dfrac{{{x^2} + 5x - 3}}{{x + 2}}} \right)dx$
Integrating the terms or the equation, we get
$\int {dy} = \int {\left( {\dfrac{{{x^2} + 5x - 3}}{{x + 2}}} \right)dx}$
As a result, solving the equation predominantly, we get
$y = \int {\left( {\dfrac{{{x^2} + 5x - 3}}{{x + 2}}} \right)dx}$
Now,
$\because$Dividing ${x^2} + 5x - 3$ by $x + 2$, we get
$y = \int {\left( {x + 3 - \dfrac{9}{{x + 2}}} \right)dx}$
Separating the above equation that is terms included in the integration sign (in this case say, integration), we get
$y = \int {\left( {x + 3} \right)dx} - 9\int {\dfrac{1}{{x + 2}}dx}$ … (i)
Let us solve the above equation separately so as to get the clarity of the solution, we get
Therefore, we will differentiate the two terms that are ‘$x + 3$’ and ‘$\dfrac{1}{{x + 2}}$’ respectively.
So, ${I_1} = \int {\left( {x + 3} \right)dx}$
Again, separating the terms inside the integration, we get
${I_1} = \int {xdx} + \int {3dx}$
${I_1} = \int {xdx} + 3\int {dx}$
As a result, solving the equation/integration by using the formula $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$ where, ‘$c$’ is (any) integration constant, we get
${I_1} = \dfrac{{{x^2}}}{2} + {c_1} + 3x + {c_2}$
Since, considering the common integration constant as ‘$c$’, we get
${I_1} = \dfrac{{{x^2}}}{2} + 3x + c$ … (ii)
Similarly,
Considering the next term of equation (i) i.e$\dfrac{1}{{x + 2}}$ one, we get.
${I_2} = \int {\dfrac{1}{{x + 2}}dx}$
Since, it is presume that the numerator is the exact derivative of denominator
Hence, we know that $\int {\left( {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}} \right)dx} = \log \left[ {f\left( x \right)} \right] + c$ where, ‘$c$’ is (any) integration constant, we get
${I_2} = \int {\dfrac{1}{{x + 2}}dx} = \log \left( {x + 2} \right) + c$ … (iii)
As a result, from equations (ii) and (iii),
Equation (i) becomes,
$y = \dfrac{{{x^{_2}}}}{2} + 3x + c - 9\left[ {\log \left( {x + 2} \right) + c} \right]$
$y = \dfrac{{{x^{_2}}}}{2} + 3x + c - 9\log \left( {x + 2} \right) - 9c = \dfrac{{{x^{_2}}}}{2} + 3x - 9\log \left( {x + 2} \right) - 8c$
Since, in this case, we will assume ‘$- 8c = C$’ as a unique constant of the solution
$y = \dfrac{{{x^{_2}}}}{2} + 3x - 9\log \left( {x + 2} \right) + C$ is the required general solution of the solution.
$\therefore$The option (c) is correct.

Note : One must be able to know the divide between the linear equations with the quadratic or tri-linear equation (as explained above in the solution). Also, one must remember the formulae of integration such as $\int {\left( {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}} \right)dx} = \log \left[ {f\left( x \right)} \right] + c$, $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$. Assume suitable integration constants (for the entire solution), so as to be sure of our final answer.