The product of variable $x$ and differential $dx$ is added to the product of square of sine of angle $y$ by $x$ and the subtraction of the product of variable $x$ and differential $dy$ from the product of variable $y$ and differential $dx$. In this differential equation problem, it is given that the sum of the terms is equal to zero.
$xdx$ $+$ $\sin^2{\Big(\dfrac{y}{x}\Big)}(ydx-xdy)$ $\,=\,$ $0$
The angle in the sine function indicates that the given differential equation is a homogeneous differential equation. So, let’s learn how to solve the given homogeneous differential equation.
The differentials are mixed with the functions in the given homogeneous differential equation. So, multiply the difference of the terms by its factor as per the distributive property of multiplication over the difference in the second term on the left hand side of the equation.
$\implies$ $xdx$ $+$ $\sin^2{\Big(\dfrac{y}{x}\Big)} \times ydx$ $-$ $\sin^2{\Big(\dfrac{y}{x}\Big)} \times xdy$ $\,=\,$ $0$
$\implies$ $xdx$ $+$ $\sin^2{\Big(\dfrac{y}{x}\Big)} \times y \times dx$ $-$ $\sin^2{\Big(\dfrac{y}{x}\Big)} \times xdy$ $\,=\,$ $0$
$\implies$ $xdx$ $+$ $y \times \sin^2{\Big(\dfrac{y}{x}\Big)} \times dx$ $-$ $\sin^2{\Big(\dfrac{y}{x}\Big)} \times xdy$ $\,=\,$ $0$
$\implies$ $xdx$ $+$ $y\sin^2{\Big(\dfrac{y}{x}\Big)}dx$ $-$ $\sin^2{\Big(\dfrac{y}{x}\Big)} \times xdy$ $\,=\,$ $0$
Now, keep the terms, which consist of the differential $dx$ on one side of the equation and move the term that contains the differential $dy$ on the other side of the equation.
$\implies$ $xdx$ $+$ $y\sin^2{\Big(\dfrac{y}{x}\Big)}dx$ $\,=\,$ $\sin^2{\Big(\dfrac{y}{x}\Big)} \times xdy$
$\implies$ $xdx$ $+$ $y\sin^2{\Big(\dfrac{y}{x}\Big)}dx$ $\,=\,$ $\sin^2{\Big(\dfrac{y}{x}\Big)} \times x \times dy$
$\implies$ $xdx$ $+$ $y\sin^2{\Big(\dfrac{y}{x}\Big)}dx$ $\,=\,$ $x \times \sin^2{\Big(\dfrac{y}{x}\Big)} \times dy$
$\implies$ $xdx$ $+$ $y\sin^2{\Big(\dfrac{y}{x}\Big)}dx$ $\,=\,$ $x\sin^2{\Big(\dfrac{y}{x}\Big)} \times dy$
$\implies$ $x\sin^2{\Big(\dfrac{y}{x}\Big)} \times dy$ $\,=\,$ $xdx$ $+$ $y\sin^2{\Big(\dfrac{y}{x}\Big)}dx$
$\implies$ $x\sin^2{\Big(\dfrac{y}{x}\Big)} \times dy$ $\,=\,$ $x \times dx$ $+$ $y\sin^2{\Big(\dfrac{y}{x}\Big)} \times dx$
Look at the differential expression on the right hand side of the equation. There is a differential $dx$ in both terms. Now, take the common differential $dx$ out from the right hand side of the expression.
$\implies$ $x\sin^2{\Big(\dfrac{y}{x}\Big)} \times dy$ $\,=\,$ $dx \times \bigg(x+y\sin^2{\Big(\dfrac{y}{x}\Big)}\bigg)$
It is time to separate the differential from the remaining functions by shifting the differentials to one side of the equation.
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{x+y\sin^2{\Big(\dfrac{y}{x}\Big)}}{x\sin^2{\Big(\dfrac{y}{x}\Big)}}$
The equation is in homogeneous form. Hence, it creates a problem for solving the differential equation and it can be resolved by replacing the dependent variable $y$ as a product of one variable and a variable $x$.
Suppose $y \,=\, vx$
Now, eliminate the dependent variable $y$ by its value in the differential equation.
$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{x+(vx)\sin^2{\Big(\dfrac{vx}{x}\Big)}}{x\sin^2{\Big(\dfrac{vx}{x}\Big)}}$
It’s time to remove the homogeneous property from the function on right hand side of the equation by simplification.
$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{x+vx\sin^2{\Big(\dfrac{v \times x}{x}\Big)}}{x\sin^2{\Big(\dfrac{v \times x}{x}\Big)}}$
$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{x+vx\sin^2{\Big(\dfrac{v \times \cancel{x}}{\cancel{x}}\Big)}}{x\sin^2{\Big(\dfrac{v \times \cancel{x}}{\cancel{x}}\Big)}}$
$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{x+vx\sin^2{v}}{x\sin^2{v}}$
$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{x+v\times x \times \sin^2{v}}{x\sin^2{v}}$
$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{x+x\times v \times \sin^2{v}}{x\sin^2{v}}$
$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{x \times 1+x\times v\sin^2{v}}{x\sin^2{v}}$
There is a common factor $x$ in the numerator and it is also there in the denominator. It indicates to us that take the common factor out from the expression in the numerator.
$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{x \times (1+v\sin^2{v})}{x \times \sin^2{v}}$
$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{\cancel{x} \times (1+v\sin^2{v})}{\cancel{x} \times \sin^2{v}}$
$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{1+v\sin^2{v}}{\sin^2{v}}$
The right hand side expression was simplified in the above step and it is required to evaluate the derivative of product of the variables. It can be done by using the product rule of differentiation.
$\implies$ $v \times \dfrac{dx}{dx}$ $+$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1+v\sin^2{v}}{\sin^2{v}}$
$\implies$ $v \times 1$ $+$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1+v\sin^2{v}}{\sin^2{v}}$
$\implies$ $v$ $+$ $x\dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1+v\sin^2{v}}{\sin^2{v}}$
Now, we have a first order and first degree differential equation and it can be solved by the separation of variables method.
$\implies$ $x\dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1+v\sin^2{v}}{\sin^2{v}}$ $-$ $v$
$\implies$ $x\dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1+v\sin^2{v}-v \times \sin^2{v}}{\sin^2{v}}$
$\implies$ $x\dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1+v\sin^2{v}-v\sin^2{v}}{\sin^2{v}}$
$\implies$ $x\dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1+\cancel{v\sin^2{v}}-\cancel{v\sin^2{v}}}{\sin^2{v}}$
$\implies$ $x\dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1}{\sin^2{v}}$
$\implies$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1}{\sin^2{v}}$
It is time to finish the process of the variable separable method.
$\implies$ $x \times dv$ $\,=\,$ $\dfrac{1}{\sin^2{v}} \times dx$
$\implies$ $\sin^2{v} \times dv$ $\,=\,$ $\dfrac{1}{x} \times dx$
$\implies$ $\sin^2{v}\,dv$ $\,=\,$ $\dfrac{1}{x}\,dx$
Now, take the indefinite integral on both sides of the equation.
$\implies$ $\displaystyle \int{\sin^2{v}}\,dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$
In integral calculus, there is no formula for finding the integral of the sine squared function but it can be calculated by reducing the power of the sine function.
$\implies$ $\displaystyle \int{\dfrac{1-\cos{(2v)}}{2}}\,dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$
$\implies$ $\displaystyle \int{\dfrac{1 \times (1-\cos{(2v)})}{2}}\,dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$
$\implies$ $\displaystyle \int{\bigg(\dfrac{1}{2} \times (1-\cos{(2v)}\bigg)}\,dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$
There is constant factor inside the integral on the left hand side of the equation. It can be excluded from the integral operation as per the constant multiple rule of integration.
$\implies$ $\dfrac{1}{2} \times \displaystyle \int{\big(1-\cos{(2v)}\big)}\,dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$
The integral of the difference of the terms can be evaluated by the difference of their integrals as per the difference rule of integration.
$\implies$ $\dfrac{1}{2} \times \bigg(\displaystyle \int{1}\,dv$ $-$ $\displaystyle \int{\cos{(2v)}}\,dv\bigg)$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$
There is no integral formula for finding the integral of the cosine of double angle. However, the issue can be solved by representing the double angle with a variable.
Assume $u = 2v$
$\implies$ $\dfrac{d}{dv}{(u)}$ $\,=\,$ $\dfrac{d}{dv}{(2v)}$
Differentiate this equation with respect to $v$.
$\implies$ $\dfrac{du}{dv}$ $\,=\,$ $\dfrac{d}{dv}{(2 \times v)}$
Use the constant multiple rule of derivatives for separating the constant factor from the differentiation as per the constant multiple rule of differentiation.
$\implies$ $\dfrac{du}{dv}$ $\,=\,$ $2 \times \dfrac{d}{dv}{(v)}$
According to the differentiation rule of the variable, the derivative of the variable with respect to same variable is equal to one.
$\implies$ $\dfrac{du}{dv}$ $\,=\,$ $2 \times 1$
$\implies$ $\dfrac{du}{dv}$ $\,=\,$ $2$
$\implies$ $du$ $\,=\,$ $2 \times dv$
$\implies$ $2 \times dv$ $\,=\,$ $du$
$\,\,\,\therefore\,\,\,\,\,\,$ $dv$ $\,=\,$ $\dfrac{du}{2}$
Now, transform the trigonometric integral function in terms of $x$ into the trigonometric integral function in terms of $u$.
$\implies$ $\dfrac{1}{2} \times \bigg(\displaystyle \int{1}\,dv$ $-$ $\displaystyle \int{\cos{(u)}}\,\Big(\dfrac{du}{2}\Big)\bigg)$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$
$\implies$ $\dfrac{1}{2} \times \bigg(\displaystyle \int{1}\,dv$ $-$ $\displaystyle \int{\cos{u}}\,\Big(\dfrac{1 \times du}{2}\Big)\bigg)$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$
$\implies$ $\dfrac{1}{2} \times \bigg(\displaystyle \int{1}\,dv$ $-$ $\displaystyle \int{\cos{u}}\,\Big(\dfrac{1}{2}\times du\Big)\bigg)$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$
$\implies$ $\dfrac{1}{2} \times \bigg(\displaystyle \int{1}\,dv$ $-$ $\displaystyle \int{\cos{u}} \times \dfrac{1}{2}\times du\bigg)$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$
$\implies$ $\dfrac{1}{2} \times \bigg(\displaystyle \int{1}\,dv$ $-$ $\displaystyle \int{\dfrac{1}{2}\times \cos{u}} \times du\bigg)$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$
Use the constant multiple rule of integration once again to separate the constant factor from the integration.
$\implies$ $\dfrac{1}{2} \times \bigg(\displaystyle \int{1}\,dv$ $-$ $\dfrac{1}{2}\times \displaystyle \int{\cos{u}} \times du\bigg)$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$
$\implies$ $\dfrac{1}{2} \times \bigg(\displaystyle \int{1}\,dv$ $-$ $\dfrac{1}{2}\times \displaystyle \int{\cos{u}}du\bigg)$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$
Use the integration rule of one to find the integral of $1$ and integration rule of cosine to find the integral of the cosine function. Similarly, use the reciprocal integration rule to find the integral of the reciprocal of variable $x$.
$\implies$ $\dfrac{1}{2} \times \bigg(v+c_1$ $-$ $\dfrac{1}{2}\times (\sin{u}+c_2)\bigg)$ $\,=\,$ $\log_{e}{x}+c_3$
In this equation, $c_1$, $c_2$ and $c_3$ are the constants of the respective integrals.
$\implies$ $\dfrac{1}{2} \times \bigg(v+c_1$ $-$ $\dfrac{1}{2}\times \sin{u}$ $-$ $\dfrac{1}{2}\times c_2\bigg)$ $\,=\,$ $\log_{e}{x}+c_3$
$\implies$ $\dfrac{1}{2} \times \bigg(v+c_1$ $-$ $\dfrac{1}{2}\,\sin{u}$ $-$ $\dfrac{1\times c_2}{2}\bigg)$ $\,=\,$ $\log_{e}{x}+c_3$
$\implies$ $\dfrac{1}{2} \times \bigg(v+c_1-\dfrac{1}{2}\,\sin{u}-\dfrac{c_2}{2}\bigg)$ $\,=\,$ $\log_{e}{x}+c_3$
Now, separate the constants to represent all constants in this equation by a constant simply.
$\implies$ $\dfrac{1}{2} \times \bigg(v-\dfrac{1}{2}\,\sin{u}+c_1-\dfrac{c_2}{2}\bigg)$ $\,=\,$ $\log_{e}{x}+c_3$
$\implies$ $\dfrac{1}{2} \times \bigg(v-\dfrac{1}{2}\,\sin{u}+c_1-\dfrac{c_2}{2}\bigg)$ $\,=\,$ $\log_{e}{x}+c_3$
$\implies$ $\dfrac{1}{2} \times \bigg(\Big(v-\dfrac{1}{2}\,\sin{u}\Big)+\Big(c_1-\dfrac{c_2}{2}\Big)\bigg)$ $\,=\,$ $\log_{e}{x}+c_3$
$\implies$ $\dfrac{1}{2} \times \Big(v-\dfrac{1}{2}\,\sin{u}\Big)$ $+$ $\dfrac{1}{2} \times \Big(c_1-\dfrac{c_2}{2}\Big)$ $\,=\,$ $\log_{e}{x}+c_3$
$\implies$ $\dfrac{1}{2}\Big(v-\dfrac{1}{2}\,\sin{u}\Big)$ $+$ $\dfrac{1}{2}\Big(c_1-\dfrac{c_2}{2}\Big)$ $\,=\,$ $\log_{e}{x}+c_3$
$\implies$ $\dfrac{1}{2}\Big(v-\dfrac{1}{2}\,\sin{u}\Big)$ $+$ $\dfrac{1}{2}\Big(c_1-\dfrac{c_2}{2}\Big)-c_3$ $\,=\,$ $\log_{e}{x}$
$\implies$ $\log_{e}{x}$ $\,=\,$ $\dfrac{1}{2}\Big(v-\dfrac{1}{2}\,\sin{u}\Big)$ $+$ $\dfrac{1}{2}\Big(c_1-\dfrac{c_2}{2}\Big)-c_3$
The second and third terms represent two constants on the right hand side expression and the difference of them is also a constant. Therefore, it can be simply denoted by a constant $c$.
$\implies$ $\log_{e}{x}$ $\,=\,$ $\dfrac{1}{2}\Big(v-\dfrac{1}{2}\,\sin{u}\Big)$ $+$ $c$
We have taken that $u \,=\, 2v$. So, replace the variable $u$ by its value in the equation.
$\implies$ $\log_{e}{x}$ $\,=\,$ $\dfrac{1}{2}\bigg(v-\dfrac{1}{2}\,\sin{(2v)}\bigg)$ $+$ $c$
The solution of the given homogeneous differential equation is obtained in terms of $v$ but the homogeneous differential equation is given in terms of $x$ and $y$. So, substitute the value of $v$ in terms of $x$ and $y$ to get the required solution.
$\implies$ $\log_{e}{x}$ $\,=\,$ $\dfrac{1}{2}\Bigg(\dfrac{y}{x}-\dfrac{1}{2}\,\sin{\bigg(2 \times \dfrac{y}{x}\bigg)}\Bigg)$ $+$ $c$
$\implies$ $\log_{e}{x}$ $\,=\,$ $\dfrac{1}{2}\Bigg(\dfrac{y}{x}-\dfrac{1}{2}\,\sin{\bigg(\dfrac{2 \times y}{x}\bigg)}\Bigg)$ $+$ $c$
$\,\,\,\therefore\,\,\,\,\,\,$ $\log_{e}{x}$ $\,=\,$ $\dfrac{1}{2}\Bigg(\dfrac{y}{x}-\dfrac{1}{2}\,\sin{\bigg(\dfrac{2y}{x}\bigg)}\Bigg)$ $+$ $c$
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