Math Doubts

Solve $e^{\displaystyle x} \tan{y}\,dx$ $+$ $(1-e^{\displaystyle x})\sec^{2}{y}\,dy$ $\,=\,$ $0$

A differential equation is given in this problem and it is formed by the combination of both exponential and trigonometric functions as follows.

$e^{\displaystyle x} \tan{y}\,dx$ $+$ $(1-e^{\displaystyle x})\sec^{2}{y}\,dy$ $\,=\,$ $0$

The product of the natural exponential function in terms of $x$, the tan of angle $y$ and the differential element $dx$ is added to the product of the substation of natural exponential function from $1$, the square of secant of angle $y$ and the differential $dy$. It is given that the sum of the both products is equal to $0$. So, the differential equation should be solved in this differential equational problem.

Separate the functions of both variables

In the given differential equation, the functions in terms of $x$ are mixed with the functions in terms of $y$. For solving the given differential equation, the functions in terms of $x$ should be shifted to one side of the equation and the functions in terms of $y$ are shifted to other side of the equation. So, let’s try to separate them for preparing the differential equation for integration as per the separation of variables method.

$\implies$ $e^{\displaystyle x} \tan{y}\,dx$ $\,=\,$ $-$ $(1-e^{\displaystyle x})\sec^{2}{y}\,dy$

$\implies$ $e^{\displaystyle x} \times \tan{y}\,dx$ $\,=\,$ $-$ $(1-e^{\displaystyle x}) \times \sec^{2}{y}\,dy$

$\implies$ $\dfrac{e^{\displaystyle x}}{1-e^{\displaystyle x}}\,dx$ $\,=\,$ $-$ $\dfrac{\sec^{2}{y}}{\tan{y}}\,dy$

Integrate both sides of the equation

Take the indefinite integration on both sides of the equation.

$\implies$ $\displaystyle \int{\dfrac{e^{\displaystyle x}}{1-e^{\displaystyle x}}}\,dx$ $\,=\,$ $-$ $\displaystyle \int{\dfrac{\sec^{2}{y}}{\tan{y}}}\,dy$

Observe the rational expressions on both sides of the equation. The expression in the numerator can be obtained by differentiating the respective expression in the denominator. Therefore, let us implement the technique of transforming the expression in terms of another variable for simplification.

$\implies$ $\displaystyle \int{\dfrac{1 \times e^{\displaystyle x}}{1-e^{\displaystyle x}}}\,dx$ $\,=\,$ $-$ $\displaystyle \int{\dfrac{1 \times \sec^{2}{y}}{\tan{y}}}\,dy$

$\implies$ $\displaystyle \int{\dfrac{1}{1-e^{\displaystyle x}}} \times e^{\displaystyle x}\,dx$ $\,=\,$ $-$ $\displaystyle \int{\dfrac{1}{\tan{y}}}\times \sec^{2}{y}\,dy$

Take $u$ $\,=\,$ $1-e^{\displaystyle x}$

Find the derivative of both sides of the equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{(u)}$ $\,=\,$ $\dfrac{d}{dx}{\big(1-e^{\displaystyle x}\big)}$

Use the difference rule of differentiation to find the derivative of difference of two terms on the right hand side of the equation.

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(1)}$ $-$ $\dfrac{d}{dx}{\big(e^{\displaystyle x}\big)}$

Use derivative of constant and derivative of natural exponential function rule to find their derivatives on the right hand side of the equation.

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $0$ $-$ $e^{\displaystyle x}$

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $-e^{\displaystyle x}$

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $\big(-e^{\displaystyle x}\big)$

$\implies$ $du$ $\,=\,$ $\big(-e^{\displaystyle x}\big) \times dx$

$\implies$ $-e^{\displaystyle x}dx$ $\,=\,$ $du$

$\,\,\,\therefore\,\,\,\,\,\,$ $e^{\displaystyle x}dx$ $\,=\,$ $-du$

Now, convert the mathematical expression on left hand side of the equation in terms $u$.

$\implies$ $\displaystyle \int{\dfrac{1}{u}} \times (-du)$ $\,=\,$ $-$ $\displaystyle \int{\dfrac{1}{\tan{y}}}\times \sec^{2}{y}\,dy$

$\implies$ $\displaystyle \int{\dfrac{1}{u}} \times (-1) \times du$ $\,=\,$ $-$ $\displaystyle \int{\dfrac{1}{\tan{y}}}\times \sec^{2}{y}\,dy$

$\implies$ $\displaystyle \int{(-1) \times \dfrac{1}{u}} \times du$ $\,=\,$ $-$ $\displaystyle \int{\dfrac{1}{\tan{y}}}\times \sec^{2}{y}\,dy$

$\implies$ $(-1) \times \displaystyle \int{\dfrac{1}{u}} \times du$ $\,=\,$ $-$ $\displaystyle \int{\dfrac{1}{\tan{y}}}\times \sec^{2}{y}\,dy$

$\implies$ $-\displaystyle \int{\dfrac{1}{u}}\,du$ $\,=\,$ $-$ $\displaystyle \int{\dfrac{1}{\tan{y}}}\times \sec^{2}{y}\,dy$

$\implies$ $\displaystyle \int{\dfrac{1}{u}}\,du$ $\,=\,$ $\displaystyle \int{\dfrac{1}{\tan{y}}}\times \sec^{2}{y}\,dy$

Assume $v$ $\,=\,$ $\tan{y}$

Now, differentiate the both sides of the equation with respect to $y$.

$\implies$ $\dfrac{d}{dy}{(v)}$ $\,=\,$ $\dfrac{d}{dy}{(\tan{y})}$

Use the derivative rule of tan function to find the differentiation of the tangent of angle $y$ with respect to $y$.

$\implies$ $\dfrac{dv}{dy}$ $\,=\,$ $\sec^2{y}$

$\implies$ $dv$ $\,=\,$ $\sec^2{y} \times dy$

$\implies$ $\sec^2{y} \times dy$ $\,=\,$ $dv$

$\,\,\,\therefore\,\,\,\,\,\,$ $dv$ $\,=\,$ $\sec^2{y} \times dy$

Now, transform the mathematical expression on right hand side of the equation in terms of $v$.

$\implies$ $\displaystyle \int{\dfrac{1}{u}}\,du$ $\,=\,$ $\displaystyle \int{\dfrac{1}{v}}\times dv$

$\implies$ $\displaystyle \int{\dfrac{1}{u}}\,du$ $\,=\,$ $\displaystyle \int{\dfrac{1}{v}}\,dv$

According to the reciprocal integral rule, the integration of the reciprocal of a variable is natural logarithm of variable.

$\implies$ $\log_{\displaystyle e}{(u)}+c_1$ $\,=\,$ $\log_{\displaystyle e}{(v)}+c_2$

Solution in terms of actual variables

The solution is obtained in terms of $u$ and $v$ but the given differential equation is given in terms of $x$ and $y$. So, replace the variables $u$ and $v$ by their actual values.

$\implies$ $\log_{\displaystyle e}{\big(1-e^{\displaystyle x}\big)}+c_1$ $\,=\,$ $\log_{\displaystyle e}{\big(\tan{y}\big)}+c_2$

$\implies$ $\log_{\displaystyle e}{\big(1-e^{\displaystyle x}\big)}$ $+$ $c_1$ $-$ $c_2$ $\,=\,$ $\log_{\displaystyle e}{\big(\tan{y}\big)}$

The difference of the constants is simply denoted by a constant $c_3$.

$\implies$ $\log_{\displaystyle e}{\big(1-e^{\displaystyle x}\big)}$ $+$ $c_3$ $\,=\,$ $\log_{\displaystyle e}{\big(\tan{y}\big)}$

The value of constant $c_3$ is denoted by the logarithm of a constant $c$ and this step is useful to eliminate the equation from the logarithm.

$\implies$ $\log_{\displaystyle e}{\big(1-e^{\displaystyle x}\big)}$ $+$ $\log{c}$ $\,=\,$ $\log_{\displaystyle e}{\big(\tan{y}\big)}$

The sum of the quantities in logarithmic form can be added by the product rule of logarithms.

$\implies$ $\log_{\displaystyle e}{\Big(\big(1-e^{\displaystyle x}\big) \times c\Big)}$ $\,=\,$ $\log_{\displaystyle e}{\big(\tan{y}\big)}$

$\implies$ $\big(1-e^{\displaystyle x}\big) \times c$ $\,=\,$ $\tan{y}$

$\implies$ $\tan{y}$ $\,=\,$ $\big(1-e^{\displaystyle x}\big) \times c$

$\implies$ $\tan{y}$ $\,=\,$ $c \times \big(1-e^{\displaystyle x}\big)$

$\,\,\,\therefore\,\,\,\,\,\,$ $\tan{y}$ $\,=\,$ $c\big(1-e^{\displaystyle x}\big)$

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