Question

The general solution of \[x\left(x-1\right)\frac{\mathrm{d}y}{\mathrm{d}x}=x^{3}\left(2x-1\right)+\left(x-2\right)y\mathrm{}\] is 

Options

• y(x - 1) = x³ + c(x - 1), where c is the constant of integration.

• y = x³(x - 1) + c, where c is the constant of integration.

• y(x - 1) = x³(x − 1) + cx², where c is the constant of integration.

• y(x - 1) = x³(x - 1) + c, where c is the constant of integration.

Answer 1

y(x - 1) = x³(x − 1) + cx², where c is the constant of integration.

Explanation:

\[x\left(x-1\right)\frac{\mathrm{d}y}{\mathrm{d}x}=x^3\left(2x-1\right)+\left(x-2\right)y\]

\[\therefore\quad x\left(x-1\right)\frac{\mathrm{d}y}{\mathrm{d}x}-\left(x-2\right)y=x^{3}\left(2x-1\right)\]

\[\therefore\quad\frac{\mathrm{d}y}{\mathrm{d}x}-\frac{x-2}{x(x-1)}y=\frac{x^3(2x-1)}{x(x-1)}\]

This is of the type

\[\frac{\mathrm{d}y}{\mathrm{d}x}+\mathrm{P}y=\mathrm{Q}\]

\[\mathrm{I.F.}=\mathrm{e}^{[\mathrm{Pd}x}\]

\[\mathrm{P}=\frac{2-x}{x(x-1)}=\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{x-1}\]

\[\therefore\quad2-x=\mathrm{A~}(x-1)+\mathrm{B}x\]

\[\therefore\quad2-x=(\mathrm{A}+\mathrm{B})x-\mathrm{A}\]

\[\therefore\quad\mathrm{A}=-2\]

B = 1

\[\mathrm{I.F.}=\mathrm{e}^{\int\mathbf{P}\mathbf{d}x}=\mathrm{e}^{-\int\frac{2}{x}\mathrm{d}x+\int\frac{1}{x-1}\mathrm{d}x}\]

\[=\mathrm{~e}^{-2\log x+\log(x-1)}\]

\[=\mathrm{~e}^{\log x^{-2}}+\log(x-1)\]

\[=\mathrm{~e}^{\log\left(\frac{x-1}{x^2}\right)}\]

\[=\frac{x-1}{x^2}\]

Solution is

\[y\left(\frac{x-1}{x^2}\right)=\int\frac{x-1}{x^2}.\frac{x^2(2x-1)}{x-1}\mathrm{d}x+\mathrm{c}\]

\[=x^2-x+\mathrm{C}\]

\[\therefore\quad y\left(x-1\right)=x^4-x^3+\mathrm{c}x^2\]

\[\therefore\quad y\left(x-1\right)=x^3\left(x-1\right)+\mathrm{c}x^2\]