Question

The solution of the differential equation $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{\text{y}}}{{\text{2}}}{\text{secx = }}\dfrac{{{\text{tanx}}}}{{{\text{2y}}}}$, where ${\text{0}} \leqslant {\text{x < }}\dfrac{{\pi }}{{\text{2}}}$ and ${\text{y}}\left( {\text{0}} \right){\text{ = 1}}$, is given by:
A) ${{\text{y}}^{\text{2}}}{\text{ = 1 + }}\dfrac{{\text{x}}}{{{\text{secx + tanx}}}}$
B) ${\text{y = 1 + }}\dfrac{{\text{x}}}{{{\text{secx + tanx}}}}$
C) ${\text{y = 1 - }}\dfrac{{\text{x}}}{{{\text{secx + tanx}}}}$
D) ${{\text{y}}^{\text{2}}}{\text{ = 1 - }}\dfrac{{\text{x}}}{{{\text{secx + tanx}}}}$

Answer 1

We can make the nonlinear differential equation to linear differential equation by dividing the equation with a suitable power of ${\text{y}}$ and then by giving a suitable substitution. We can solve the linear ordinary differential equation by finding the integrating factor and integrating. Then apply the initial value to get the value of the constant of integration.

Complete step by step solution: We have a nonlinear ordinary differential equation.
$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{\text{y}}}{{\text{2}}}{\text{secx = }}\dfrac{{{\text{tanx}}}}{{{\text{2y}}}}$
We can make the equation by following steps. Firstly, we can divide the equation by $\dfrac{{\text{1}}}{{\text{y}}}$,
$\dfrac{{{\text{ydy}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{{\text{y}}^{\text{2}}}}}{{\text{2}}}{\text{secx = }}\dfrac{{{\text{tanx}}}}{{\text{2}}}$
Let ${\text{v = }}{{\text{y}}^{\text{2}}}$,then,
$\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ = 2y}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$
$\Rightarrow {\text{y}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}$
Making this substitution in , we get,
$\dfrac{{\text{1}}}{{\text{2}}}{ \times }\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{\text{v}}}{{\text{2}}}{\text{secx = }}\dfrac{{{\text{tanx}}}}{{\text{2}}}$
Multiplying the equation throughout with 2, we get,
$\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ + vsecx = tanx}}$
Now we have a linear differential equation with ${\text{P}}\left( {\text{x}} \right){\text{ = secx}}$ and ${\text{Q}}\left( {\text{x}} \right){\text{ = tanx}}$. Then the integrating factor is given by,
${\text{IF = }}{{\text{e}}^{\int {{\text{P}}\left( {\text{x}} \right){\text{dx}}} }}{\text{ = }}{{\text{e}}^{\int {{\text{secxdx}}} }}$
We know that, $\int {{\text{secxdx = ln}}\left| {{\text{tanx + secx}}} \right|}$,
$\Rightarrow {\text{IF = }}{{\text{e}}^{{\text{ln}}\left| {{\text{tanx + secx}}} \right|}}$
We know that, ${{\text{e}}^{{\text{ln}}\left| {\text{a}} \right|}}{\text{ = a}}$. So, we get,
$\Rightarrow {\text{IF = tanx + secx}}$
Now we can solve for ${\text{v}}$,
${\text{v = }}\dfrac{{\text{1}}}{{{\text{IF}}}}\int {{\text{IF} \times \text{Q}}\left( {\text{x}} \right){\text{dx}}} {\text{ + C = }}\dfrac{{\text{1}}}{{{\text{tanx + secx}}}}\int {{\text{tanx(tanx + secx)dx}}} {\text{ + C}}$
We can solve the integration part.
$\int {{\text{tanx}}\left( {{\text{tanx + secx}}} \right){\text{dx}}} {\text{ = }}\int {{\text{(ta}}{{\text{n}}^{\text{2}}}{\text{x + tanxsecx)dx}}}$
We know that ${\text{ta}}{{\text{n}}^{\text{2}}}{\text{x = se}}{{\text{c}}^{\text{2}}}{\text{x - 1}}$. So, we get,
${\text{ = }}\int {\left( {{\text{se}}{{\text{c}}^{\text{2}}}{\text{x - 1 + tanxsecx}}} \right)} {\text{dx}} \\\ {\text{ = }}\int {{\text{se}}{{\text{c}}^{\text{2}}}{\text{xdx}}} {\text{ - }}\int {{\text{1dx}}} {\text{ + }}\int {{\text{tanxsecxdx}}} \\\$
We know the integral of $\int {{\text{se}}{{\text{c}}^{\text{2}}}{\text{xdx}}} {\text{ = tanx}}$, $\int {{\text{1dx}}} {\text{ = x}}$and $\int {{\text{tanxsecxdx}}} {\text{ = secx}}$. So, we get,
$\int {{\text{tanx}}\left( {{\text{tanx + secx}}} \right){\text{dx}}} {\text{ = tanx - x + secx}}$
So, ${\text{v}}$becomes,
${\text{v = }}\dfrac{{{\text{tanx + secx - x}}}}{{{\text{tanx + secx}}}}{\text{ + C}}$
Resubstituting ${\text{v = }}{{\text{y}}^{\text{2}}}$, we get,
${{\text{y}}^{\text{2}}}{\text{ = 1 - }}\dfrac{{\text{x}}}{{{\text{tanx + secx}}}}{\text{ + C}}$
Applying initial condition, ${\text{y}}\left( {\text{0}} \right){\text{ = 1}}$,

$${\text{1 = 1 - }}\dfrac{{\text{0}}}{{{\text{tanx + secx}}}}{\text{ + C}} \\\ \Rightarrow {\text{C = 1 - 1 = 0}} \\\$$

Giving, ${\text{C = 0}}$, we get,
${{\text{y}}^{\text{2}}}{\text{ = 1 - }}\dfrac{{\text{x}}}{{{\text{tanx + secx}}}}$

Therefore, the correct answer is option D.

Note: A non-linear ordinary differential equation of the form $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + yP}}\left( {\text{x}} \right){\text{ = }}{{\text{y}}^{\text{n}}}{\text{Q}}\left( {\text{x}} \right)$ can be converted into linear form by dividing the equation by ${{\text{y}}^{\text{n}}}$, the give substitution for ${{\text{y}}^{{\text{1 - n}}}}$and divide by ${\text{1 - n}}$.
We can solve the linear differential equation by finding the integrating factor and integrating. As the initial value is given, we must find the value of the constant of integration by applying the initial condition. This type of problem with the initial values given is called initial value problems.