Solve $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{y}{x}$ $+$ $\sin{\Big(\dfrac{y}{x}\Big)}$

The derivative of $y$ with respect to $x$ is equal to the sum of the quotient of $y$ by $x$ and sine of angle quotient of $y$ by $x$ in this differential equation problem.

$\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{y}{x}$ $+$ $\sin{\Big(\dfrac{y}{x}\Big)}$

The above differential equation represents the homogeneous differential equation. So, let’s learn how to solve this homogeneous differential equation mathematically with understandable steps.

Eliminate a variable by a suitable value

The involvement of $x$ and $y$ in quotient form made the differential equation complex. For eliminating the complexity of the differential equation, let’s represent the product of two variables by a variable.

Suppose $y \,=\, vx$ and eliminate the complexity of the differential equation by substitution.

$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{vx}{x}$ $+$ $\sin{\Big(\dfrac{vx}{x}\Big)}$

It is time to simplify the mathematical expression on right hand side of the equation.

$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{v \times x}{x}$ $+$ $\sin{\Big(\dfrac{v \times x}{x}\Big)}$

$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{v \times \cancel{x}}{\cancel{x}}$ $+$ $\sin{\Big(\dfrac{v \times \cancel{x}}{\cancel{x}}\Big)}$

$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $v$ $+$ $\sin{(v)}$

Evaluate the derivative of the product

It is right time to calculate the derivative of the product of the variable on left-hand side of the equation. It can be done by using the product rule of the derivatives.

$\implies$ $v \times \dfrac{dx}{dx}$ $+$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $v$ $+$ $\sin{v}$

$\implies$ $v \times 1$ $+$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $v$ $+$ $\sin{v}$

$\implies$ $v$ $+$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $v$ $+$ $\sin{v}$

Solve the equation by variable separable

The homogeneous form is successfully eliminated from the differential equation. Hence, it can be solved by using the separation of variables method.

$\implies$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $v$ $-$ $v$ $+$ $\sin{v}$

$\implies$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $\cancel{v}$ $-$ $\cancel{v}$ $+$ $\sin{v}$

$\implies$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $\sin{v}$

$\implies$ $x \times dv$ $\,=\,$ $\sin{v} \times dx$

$\implies$ $\dfrac{dv}{\sin{v}}$ $\,=\,$ $\dfrac{dx}{x}$

$\implies$ $\dfrac{1 \times dv}{\sin{v}}$ $\,=\,$ $\dfrac{1 \times dx}{x}$

$\implies$ $\dfrac{1}{\sin{v}}\times dv$ $\,=\,$ $\dfrac{1}{x} \times dx$

$\implies$ $\dfrac{1}{\sin{v}}\,dv$ $\,=\,$ $\dfrac{1}{x}\,dx$

According to the reciprocal identity of sine function, the reciprocal of sine function can be written as cosecant.

$\implies$ $\csc{v}\,dv$ $\,=\,$ $\dfrac{1}{x}\,dx$

Take the indefinite integrals on both sides of the equation for solving this differential equation mathematically.

$\implies$ $\displaystyle \int{\csc{v}}\,dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$

Find the integral of cosecant function and also find the reciprocal of variable by using the reciprocal rule of integrals.

$\implies$ $\log_{e}{(\csc{v}-\cot{v})}$ $+$ $c_1$ $\,=\,$ $\log_{e}{(x)}$ $+$ $c_2$

$\implies$ $\log_{e}{(\csc{v}-\cot{v})}$ $-$ $\log_{e}{(x)}$ $\,=\,$ $c_2-c_1$

The subtraction of the logarithms can be calculated by using the quotient rule of logarithms. Similarly, the difference of the constants can be simply denoted by another constant $c_3$.

$\implies$ $\log_{e}{\bigg(\dfrac{\csc{v}-\cot{v}}{x}\bigg)}$ $\,=\,$ $c_3$

The expression on the left-hand side of the equation can be released from the logarithms by using the mathematical relationship between the logarithms and exponentials.

$\implies$ $\dfrac{\csc{v}-\cot{v}}{x}$ $\,=\,$ $e^{\displaystyle c_3}$

The expression on the right-hand side of the equation is a constant. So, it can be simply denoted by a constant $c$.

$\implies$ $\dfrac{\csc{v}-\cot{v}}{x}$ $\,=\,$ $c$

Now, simplify the mathematical equation further to get the solution of the differential equation.

$\implies$ $\csc{v}-\cot{v}$ $\,=\,$ $c \times x$

$\implies$ $\csc{v}-\cot{v}$ $\,=\,$ $cx$

The solution of the differential equation is obtained in terms of the variables $x$ and $v$ but the differential equation is given in terms of $x$ and $y$. Hence, replace the value of $v$ in terms of $x$ and $y$ to get the required solution of the given differential equation.

$\,\,\,\therefore\,\,\,\,\,\,$ $\csc{\Big(\dfrac{y}{x}\Big)}$ $-$ $\cot{\Big(\dfrac{y}{x}\Big)}$ $\,=\,$ $cx$