Question

(y+x^2/2+y^3/3)dx+ 1/4(x+xy^2)dy=0​

Answer 1

Given differential equation is,

$\longrightarrow \left(y+\dfrac{x^2}{2}+\dfrac{y^3}{3}\right)\,dx+\dfrac{1}{4}\left(x+xy^2\right)\,dy=0$

Multiply each term by $x^my^n.$

$\footnotesize\longrightarrow\left(x^my^{n+1}+\dfrac{1}{2}\,x^{m+2}y^n+\dfrac{1}{3}\,x^my^{n+3}\right)\,dx+\dfrac{1}{4}\left(x^{m+1}y^n+x^{m+1}y^{n+2}\right)\,dy=0\dots(1)$

Let the solution of this differential equation be in the form,

$\longrightarrow S(x,\ y)=C\quad\quad\dots(i)$

where $S$ is a function in $x$ and $y,$ such that,

$\longrightarrow\dfrac{\partial S}{\partial x}=\dfrac{\partial S}{\partial y}=0$

When we find $\dfrac{\partial S}{\partial x}$ and $\dfrac{\partial S}{\partial y}$ without multiplying the equation by $x^my^n$ the condition $\dfrac{\partial S}{\partial x}=\dfrac{\partial S}{\partial y}$ may not be satisfied, that's why $x^my^n$ is multiplied so that the condition would be satisfied for any value of $m$ and $n.$

Assume the given differential equation be in the form,

$\longrightarrow\dfrac{\partial S}{\partial x}\,dx+\dfrac{\partial S}{\partial y}\,dy=0\quad\quad\dots(2)$

By assuming this way, the coefficients of $dx$ and $dy$ become zero each, and then LHS becomes equal to RHS that is zero.

One can also assume the equation be in the form,

$\longrightarrow\dfrac{\partial S}{\partial y}\,dx+\dfrac{\partial S}{\partial x}\,dy=0$

Comparing (1) and (2) we get,

$\longrightarrow\dfrac{\partial S}{\partial x}=x^my^{n+1}+\dfrac{1}{2}\,x^{m+2}y^n+\dfrac{1}{3}\,x^my^{n+3}\quad\quad\dots(3)$

Performing partial differentiation wrt $y,$

$\longrightarrow\dfrac{\partial^2S}{\partial x\,\partial y}=(n+1)x^my^n+\dfrac{1}{2}\,nx^{m+2}y^{n-1}+\dfrac{1}{3}\,(n+3)x^my^{n+2}\quad\quad\dots(4)$

Also,

$\longrightarrow\dfrac{\partial S}{\partial y}=\dfrac{1}{4}\left(x^{m+1}y^n+x^{m+1}y^{n+2}\right)$

$\longrightarrow\dfrac{\partial S}{\partial y}=\dfrac{1}{4}\,x^{m+1}y^n+\dfrac{1}{4}\,x^{m+1}y^{n+2}\quad\quad\dots(5)$

Performing partial differentiation wrt $x,$

$\longrightarrow\dfrac{\partial^2S}{\partial x\,\partial y}=\dfrac{1}{4}\,(m+1)x^my^n+\dfrac{1}{4}\,(m+1)x^my^{n+2}\quad\quad\dots(6)$

Equating (4) and (6),

$\footnotesize\longrightarrow(n+1)x^my^n+\dfrac{1}{2}\,nx^{m+2}y^{n-1}+\dfrac{1}{3}\,(n+3)x^my^{n+2}=\dfrac{1}{4}\,(m+1)x^my^n+\dfrac{1}{4}\,(m+1)x^my^{n+2}$

$\longrightarrow(n+1)y+\dfrac{1}{2}\,nx^2+\dfrac{1}{3}\,(n+3)y^3=\dfrac{1}{4}\,(m+1)y+\dfrac{1}{4}\,(m+1)y^3$

$\longrightarrow\left(n+1-\dfrac{m+1}{4}\right)y+\dfrac{1}{2}\,nx^2+\left(\dfrac{n+3}{3}-\dfrac{m+1}{4}\right)y^3=0$

We equate each coefficient to zero. So,

• $n+1-\dfrac{m+1}{4}=0$

• $\dfrac{1}{2}\,n=0$

• $\dfrac{n+3}{3}-\dfrac{m+1}{4}=0$

Solving them we get,

• $m=3$

• $n=0$

Then (3) becomes,

$\longrightarrow\dfrac{\partial S}{\partial x}=x^3y+\dfrac{1}{2}\,x^5+\dfrac{1}{3}\,x^3y^3$

$\longrightarrow\partial S=\left(x^3y+\dfrac{1}{2}\,x^5+\dfrac{1}{3}\,x^3y^3\right)\ \partial x$

Integrating, (note that y is treated as constant here)

$\displaystyle\longrightarrow S=\int\left(x^3y+\dfrac{1}{2}\,x^5+\dfrac{1}{3}\,x^3y^3\right)\,\partial x$

$\displaystyle\longrightarrow S=\dfrac{1}{4}\,x^4y+\dfrac{1}{12}\,x^6+\dfrac{1}{12}\,x^4y^3+f_1(y)\quad\quad\dots(7)$

where $f_1(y)$ is an arbitrary function in $y$ but is treated as constant of integration as the partial integration is done wrt $x.$

And (5) becomes,

$\longrightarrow\dfrac{\partial S}{\partial y}=\dfrac{1}{4}\,x^4+\dfrac{1}{4}\,x^4y^2$

$\longrightarrow\partial S=\left(\dfrac{1}{4}\,x^4+\dfrac{1}{4}\,x^4y^2\right)\,\partial y$

Integrating, (note that x is treated as constant here)

$\displaystyle\longrightarrow S=\int\left(\dfrac{1}{4}\,x^4+\dfrac{1}{4}\,x^4y^2\right)\,\partial y$

$\displaystyle\longrightarrow S=\dfrac{1}{4}\,x^4y+\dfrac{1}{12}\,x^4y^3+f_2(x)\quad\quad\dots(8)$

where $f_2(x)$ is similar to $f_1(y).$

Comparing (7) and (8) we get,

• $f_1(y)=0$

• $f_2(x)=\dfrac{1}{12}\,x^6$

Then,

$\displaystyle\longrightarrow S=\dfrac{1}{4}\,x^4y+\dfrac{1}{12}\,x^4y^3+\dfrac{1}{12}\,x^6$

Hence the solution to our differential equation is given by (i) as,

$\displaystyle\longrightarrow\dfrac{1}{4}\,x^4y+\dfrac{1}{12}\,x^4y^3+\dfrac{1}{12}\,x^6=C_1$

Dividing by $x^4,$

$\displaystyle\longrightarrow\dfrac{1}{4}\,y+\dfrac{1}{12}\,y^3+\dfrac{1}{12}\,x^2=\dfrac{C_1}{x^4}$

Multiplying by 12,

$\displaystyle\longrightarrow3y+y^3+x^2=\dfrac{12C_1}{x^4}$

$\displaystyle\longrightarrow y^3+3y=\dfrac{12C_1}{x^4}-x^2$

Taking $12C_1=C,$

$\displaystyle\longrightarrow\underline{\underline{y^3+3y=\dfrac{C}{x^4}-x^2}}$