# Solve $(x-y)^2\dfrac{dy}{dx}$ $\,=\,$ $a^2$

A differential equation of first order and first degree is given in terms of variables $x$ and $y$ in this differential equation problem.

$(x-y)^2\dfrac{dy}{dx}$ $\,=\,$ $a^2$

In this differential equation, the $x$ minus $y$ whole square is multiplied by the derivative of variable $y$ with respect to $x$ on the left hand side of the equation. It is given that their product is equals to $a$ squared and it’s also given in this problem that we have to solve this differential equation.

### Simplify the complexity by transformation

The factor $x-y$ whole square makes the given differential equation complicated and it creates an issue while separating the variables. Hence, a transformation technique should be used for representing the expression $x-y$ by a variable.

Take $z \,=\, x-y$

Now, differentiate the above algebraic equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{(z)}$ $\,=\,$ $\dfrac{d}{dx}{(x-y)}$

The derivative of difference between two variables can be calculated by the difference of their derivatives as per the subtraction rule of the differentiation.

$\implies$ $\dfrac{dz}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(x)}$ $-$ $\dfrac{d}{dx}{(y)}$

$\implies$ $\dfrac{dz}{dx}$ $\,=\,$ $\dfrac{dx}{dx}$ $-$ $\dfrac{dy}{dx}$

According to the derivative rule of a variable, the derivative of $x$ with respect to $x$ is equal to one.

$\implies$ $\dfrac{dz}{dx}$ $\,=\,$ $1$ $-$ $\dfrac{dy}{dx}$

Now, find the value of derivative of $y$ with respect to $x$ by simplifying the above equation.

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{dy}{dx}$ $\,=\,$ $1$ $-$ $\dfrac{dz}{dx}$

It is time to focus on the given differential equation.

$(x-y)^2\dfrac{dy}{dx}$ $\,=\,$ $a^2$

Now, convert this differential equation in terms of $z$.

$\implies$ $(z)^2\bigg(1-\dfrac{dz}{dx}\bigg)$ $\,=\,$ $a^2$

### Use the separation of variables technique

The differential equation can be prepared now for the indefinite integration by the separation of variables method.

$\implies$ $z^2 \times \bigg(1-\dfrac{dz}{dx}\bigg)$ $\,=\,$ $a^2$

$\implies$ $z^2 \times 1$ $-$ $z^2 \times \dfrac{dz}{dx}$ $\,=\,$ $a^2$

$\implies$ $z^2$ $-$ $z^2\dfrac{dz}{dx}$ $\,=\,$ $a^2$

$\implies$ $z^2$ $-$ $a^2$ $\,=\,$ $z^2\dfrac{dz}{dx}$

$\implies$ $z^2\dfrac{dz}{dx}$ $\,=\,$ $z^2-a^2$

$\implies$ $\dfrac{dz}{dx}$ $\,=\,$ $\dfrac{z^2-a^2}{z^2}$

It is time to separate the variables with its corresponding differential from another variable and its differential as per the variables separable method.

$\implies$ $\dfrac{z^2}{z^2-a^2}\,dz$ $\,=\,$ $dx$

### Solve the differential equation by integration

Take the indefinite integration on both sides of the equation.

$\implies$ $\displaystyle \int{\dfrac{z^2}{z^2-a^2}}\,dz$ $\,=\,$ $\displaystyle \int{}\,dx$

The degrees of expressions in both numerator and denominator are same.so, add $a$ squared to the numerator and subtract same quantity from the sum.

$\implies$ $\displaystyle \int{\dfrac{z^2+a^2-a^2}{z^2-a^2}}\,dz$ $\,=\,$ $\displaystyle \int{}\,dx$

$\implies$ $\displaystyle \int{\dfrac{z^2-a^2+a^2}{z^2-a^2}}\,dz$ $\,=\,$ $\displaystyle \int{}\,dx$

The expression in fraction form on left hand side of the equation can be written as sum of two fractions for our convenience.

$\implies$ $\displaystyle \int{\bigg(\dfrac{z^2-a^2}{z^2-a^2}+\dfrac{a^2}{z^2-a^2}\bigg)}\,dz$ $\,=\,$ $\displaystyle \int{}\,dx$

$\implies$ $\displaystyle \int{\bigg(\dfrac{\cancel{z^2-a^2}}{\cancel{z^2-a^2}}+\dfrac{a^2}{z^2-a^2}\bigg)}\,dz$ $\,=\,$ $\displaystyle \int{}\,dx$

$\implies$ $\displaystyle \int{\bigg(1+\dfrac{a^2}{z^2-a^2}\bigg)}\,dz$ $\,=\,$ $\displaystyle \int{}\,dx$

The integral of sum of two functions can be evaluated by the sum their integrals as per the addition rule of integrals.

$\implies$ $\displaystyle \int{(1)}dz$ $+$ $\displaystyle \int{\dfrac{a^2}{z^2-a^2}}\,dz$ $\,=\,$ $\displaystyle \int{}dx$

$\implies$ $\displaystyle \int{(1)}dz$ $+$ $\displaystyle \int{\dfrac{a^2 \times 1}{z^2-a^2}}\,dz$ $\,=\,$ $\displaystyle \int{}dx$

Separate the constant factor from the integration in the second term as per the constant multiple property of the integration.

$\implies$ $\displaystyle \int{(1)}dz$ $+$ $a^2 \times \displaystyle \int{\dfrac{1}{z^2-a^2}}\,dz$ $\,=\,$ $\displaystyle \int{}dx$

$\implies$ $\displaystyle \int{(1)}dz$ $+$ $a^2\displaystyle \int{\dfrac{1}{z^2-a^2}}\,dz$ $\,=\,$ $\displaystyle \int{(1)}dx$

According the integral rule of one, the integral of $1$ with respect to any variable is equal to respective variable. The integral of function in the second term can be calculated by the reciprocal rule of integration for the difference of squares.

$\implies$ $z+c_1$ $+$ $a^2\Bigg(\dfrac{1}{2a}\log_{e}{\Bigg|\dfrac{z-a}{z+a}\Bigg|}+c_2\Bigg)$ $\,=\,$ $x+c_3$

The solution is calculated in terms of the variables $x$ and $z$ but the given differential equation is given in terms of $x$ and $y$. hence, eliminate the variable $z$ by its equivalent value.

$\implies$ $x$ $-$ $y$ $+$ $c_1$ $+$ $a^2\Bigg(\dfrac{1}{2a}\log_{e}{\Bigg|\dfrac{x-y-a}{x-y+a}\Bigg|}+c_2\Bigg)$ $\,=\,$ $x$ $+$ $c_3$

It is time to simplify the algebraic equation to find the solutions for the given differential equation.

$\implies$ $x$ $-$ $y$ $+$ $c_1$ $+$ $a^2 \times \dfrac{1}{2a}\log_{e}{\Bigg|\dfrac{x-y-a}{x-y+a}\Bigg|}$ $+$ $a^2 \times c_2$ $\,=\,$ $x$ $+$ $c_3$

$\implies$ $x$ $-$ $y$ $+$ $c_1$ $+$ $\dfrac{a^2 \times 1}{2a}\log_{e}{\Bigg|\dfrac{x-y-a}{x-y+a}\Bigg|}$ $+$ $a^2c_2$ $\,=\,$ $x$ $+$ $c_3$

$\implies$ $x$ $-$ $y$ $+$ $c_1$ $+$ $\dfrac{a^2}{2a}\log_{e}{\Bigg|\dfrac{x-y-a}{x-y+a}\Bigg|}$ $+$ $a^2c_2$ $\,=\,$ $x$ $+$ $c_3$

$\implies$ $x$ $-$ $y$ $+$ $c_1$ $+$ $\dfrac{\cancel{a^2}}{2\cancel{a}}\log_{e}{\Bigg|\dfrac{x-y-a}{x-y+a}\Bigg|}$ $+$ $a^2c_2$ $\,=\,$ $x$ $+$ $c_3$

$\implies$ $x$ $-$ $y$ $+$ $c_1$ $+$ $\dfrac{a}{2}\log_{e}{\Bigg|\dfrac{x-y-a}{x-y+a}\Bigg|}$ $+$ $a^2c_2$ $\,=\,$ $x$ $+$ $c_3$

$\implies$ $\dfrac{a}{2}\log_{e}{\Bigg|\dfrac{x-y-a}{x-y+a}\Bigg|}$ $\,=\,$ $x$ $+$ $c_3$ $-$ $x$ $+$ $y$ $-$ $c_1$ $-$ $a^2c_2$

$\implies$ $\dfrac{a}{2}\log_{e}{\Bigg|\dfrac{x-y-a}{x-y+a}\Bigg|}$ $\,=\,$ $x$ $-$ $x$ $+$ $y$ $-$ $c_1$ $-$ $a^2c_2$ $+$ $c_3$

$\implies$ $\dfrac{a}{2}\log_{e}{\Bigg|\dfrac{x-y-a}{x-y+a}\Bigg|}$ $\,=\,$ $\cancel{x}$ $-$ $\cancel{x}$ $+$ $y$ $-$ $c_1$ $-$ $a^2c_2$ $+$ $c_3$

$\implies$ $\dfrac{a}{2}\log_{e}{\Bigg|\dfrac{x-y-a}{x-y+a}\Bigg|}$ $\,=\,$ $y$ $-$ $c_1$ $-$ $a^2c_2$ $+$ $c_3$

$\implies$ $a\log_{e}{\Bigg|\dfrac{x-y-a}{x-y+a}\Bigg|}$ $\,=\,$ $2\big(y$ $-$ $c_1$ $-$ $a^2c_2$ $+$ $c_3\big)$

$\implies$ $a\log_{e}{\Bigg|\dfrac{x-y-a}{x-y+a}\Bigg|}$ $\,=\,$ $2\big(y$ $+$ $(-$ $c_1$ $-$ $a^2c_2$ $+$ $c_3)\big)$

$\implies$ $a\log_{e}{\Bigg|\dfrac{x-y-a}{x-y+a}\Bigg|}$ $\,=\,$ $2y$ $+$ $2(-$ $c_1$ $-$ $a^2c_2$ $+$ $c_3)$

On the right hand side of the equation, the second term is a constant. Hence. It is simply denoted by a constant $c$.

$\,\,\,\therefore\,\,\,\,\,\,$ $a\log_{e}{\Bigg|\dfrac{x-y-a}{x-y+a}\Bigg|}$ $\,=\,$ $2y$ $+$ $c$

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