Question

Find the particular solution of the equation $x\left( {{x}^{2}}-1 \right)\dfrac{dy}{dx}=1$ with initial condition $y=0$ when $x=2$.

Answer 1

First we need to find the general solution of the given differential equation. We proceed by separating the differentials and the variables to different sides of the equation. We shall use a partial fraction method to break the fraction at the side involving only variables. Then we shall integrate both sides of the equation to obtain the general solution. Finally we put the given initial to obtain a particular solution.

Complete step by step answer:
From the given differential equation, $\begin{aligned} & x\left( {{x}^{2}}-1 \right)\dfrac{dy}{dx}=1 \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{x\left( {{x}^{2}}-1 \right)}=\dfrac{1}{x\left( x-1 \right)\left( x+1 \right)}.....(1) \end{aligned}$
Now we shall break the fraction into three different parts. We use some real numbers $A,B$ and $C$ to apply partial fraction methods .

& \dfrac{1}{x\left( x-1 \right)\left( x+1 \right)}=\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{x+1}. \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{x+1}....(2) \\\ \end{aligned}$$ We shall multiply $x\left( x-1 \right)\left( x+1 \right)$ to eliminate denominators from both sides,$$$$ $1=A\left( x-1 \right)\left( x+1 \right)+Bx\left( x+1 \right)+Cx\left( x-1 \right)....(3)$$$$$ Putting $x=0$ in equation (3), we get $ 1=A\left( -1 \right)\left( 1 \right)\Rightarrow A=-1 $ $$$$ Putting $x=1$ in equation (3), we get $1=B\cdot 1\cdot \left( 1+1 \right)\Rightarrow B=\dfrac{1}{2}$$$$$ Putting $x=-1$ in equation (3), we get $1=C\cdot 1\cdot \left( 1-\left( -1 \right) \right)\Rightarrow C=\dfrac{1}{2}$$$$$ Substituting the values of $A,B$ and $C$ in the equation (2) $$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{x+1}. \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{x}+\dfrac{1}{2\left( x-1 \right)}+\dfrac{1}{2\left( x+1 \right)}. \\\ \end{aligned}$$ Now we shall integrate both sides with respect to $x$, $$\begin{gathered} \int {\dfrac{{dy}}{{dx}}} dx = \int {\left[ {\dfrac{{ - 1}}{x} + \dfrac{1}{{2\left( {x - 1} \right)}} + \dfrac{1}{{2\left( {x + 1} \right)}}} \right]dx} \\\ \Rightarrow y = \int {\dfrac{{ - 1}}{x}dx + \int {\dfrac{1}{{2\left( {x - 1} \right)}}} dx + \int {\dfrac{1}{{2\left( {x + 1} \right)}}} dx} \\\ \end{gathered} $$ Using the formula $\int{\dfrac{1}{x}}dx=\log \left| x \right|+c$ where c is any real constant we get, $$\begin{aligned} & y=-\log \left| x \right|+\dfrac{1}{2}\log \left| x+1 \right|+\left( \dfrac{1}{2} \right)\log \left| x-1 \right|+C \\\ & =-\dfrac{2}{2}\log \left| x \right|+\dfrac{1}{2}\log \left| x+1 \right|+\left( \dfrac{1}{2} \right)\log \left| x-1 \right|+C \\\ \end{aligned}$$ where C is a real constant. Now we use the logarithm formulas $m{{\log }_{e}}x={{\log }_{e}}{{x}^{m}}$ and ${{\log }_{e}}\left( ab \right)={{\log }_{e}}a+{{\log }_{e}}b$ . Then we also use modulus formulas $ \left| {{a}^{m}} \right|={{\left| a \right|}^{m}}$ and $\left| a\cdot b \right|=\left| a \right|\cdot \left| b \right|$ where $m$ and $n$ are integers. $$\begin{aligned} & y=\dfrac{1}{2}\left[ \log {{\left| x \right|}^{-2}}+\log \left| \left( x-1 \right)\left( x+1 \right) \right| \right]+C \\\ & \Rightarrow y=\dfrac{1}{2}\left[ \log \left| {{x}^{-2}} \right|+\log \left| \left( {{x}^{2}}-1 \right) \right| \right]+C \\\ & \Rightarrow y=\dfrac{1}{2}\log \left| \dfrac{{{x}^{2}}-1}{{{x}^{2}}} \right|+C.....(3) \\\ \end{aligned}$$ The above equation is the general solution of the given differential equation. We were also given the initial condition that when $x=2,y=0$. Putting the given initial condition in equation (3) $$\begin{aligned} & 0=\dfrac{1}{2}\log \left| \dfrac{{{2}^{2}}-1}{{{2}^{2}}} \right|+C \\\ & \Rightarrow C=-\dfrac{1}{2}\log \dfrac{3}{4} \\\ \end{aligned}$$ So the value of the real constant is found to be $$C=-\dfrac{1}{2}\log \dfrac{3}{4}$$. Putting the value of the constant in equation (3), $$y=\dfrac{1}{2}\log \left| \dfrac{{{x}^{2}}-1}{{{x}^{2}}} \right|-\dfrac{1}{2}\log \dfrac{3}{4}....(4)$$ **Equation (4) is the required particular solution of the given differential equation.** **Note:** The method of solution differential equations by separation of variables with partial fraction method may consume time more than usual. The substation of values and usage of formulas in a careful manner can help us get the solution in simplified form.