Question

Solution of the differential equation $(x-1)dy+ydx=x(x-1)y^{\frac{1}{3}}dx,x>1$ is:
(where $c$ is integration constant)

• A. $y^{\frac{2}{3}}=\frac{1}{4}(x-1{)}^2+\frac{2}{5}(x-1)+c(x-1{)}^{-\frac{2}{3}}$
• B. $y^{\frac{2}{3}}=\frac{1}{4}(x-1{)}^2+\frac{3}{5}(x-1)+c(x-1{)}^{-\frac{2}{3}}$
• C. $y^{\frac{1}{3}}=\frac{1}{4}(x-1{)}^3+\frac{2}{5}(x-1)+c(x-1{)}^{-\frac{2}{3}}$
• D. $y=\frac{1}{4}(x-1{)}^2+\frac{2}{5}(x-1)+c(x-1{)}^{-\frac{2}{3}}$

Answer 1

The correct option is A $y^{\frac{2}{3}}=\frac{1}{4}(x-1{)}^2+\frac{2}{5}(x-1)+c(x-1{)}^{-\frac{2}{3}}$

$(x-1)dy+ydx=x(x-1)y^{\frac{1}{3}}dx$
Dividing by $dx{y}^{\frac{1}{3}}(x-1),$ the given equation reduces to
$y^{-\frac{1}{3}}\frac{dy}{dx}+\frac{1}{x-1}{y}^{\frac{2}{3}}=x$
Put $y^{\frac{2}{3}}=z,$ so that $\frac{2}{3}{y}^{-\frac{1}{3}}\frac{dy}{dx}=\frac{dz}{dx}$
then given equation reduces to
$\frac{dz}{dx}+\frac{2}{3(x-1)}z=\frac{2}{3}x$(linear form)
On comparing with,$\frac{dz}{dx}+Pz=Q,$
We get, $P=\frac{2}{3(x-1)}$&$Q=\frac{2x}{3}$

$I.F.=e^{\frac{2}{3}\int \frac{1}{x-1}dx}=e^{\frac{2}{3}\ln |x-1|}=(x-1{)}^{\frac{2}{3}}$

$\therefore$ Solution is given by
$z(x-1{)}^{\frac{2}{3}}=\frac{2}{3}\int x(x-1{)}^{\frac{2}{3}}dx+c$
Putting $(x-1)=t^3$ in the R.H.S., we get
$z(x-1{)}^{\frac{2}{3}}=\frac{2}{3}\int x(x-1{)}^{\frac{2}{3}}dx=\frac{2}{3}\int (t^3+1)t^{2}3t^{2}dt$
$z(x-1{)}^{\frac{2}{3}}=2\int (t^7+t^4)dt=2[\left(\frac{1}{8}\right)t^8+\left(\frac{1}{5}\right)t^5]+c$
$z(x-1{)}^{\frac{2}{3}}=\left(\frac{2}{8}\right)(x-1{)}^{\frac{8}{3}}+\left(\frac{2}{5}\right)(x-1{)}^{\frac{5}{3}}+c$
Hence, the solution is $y^{\frac{2}{3}}=\frac{1}{4}(x-1{)}^2+\frac{2}{5}(x-1)+c(x-1{)}^{-\frac{2}{3}}$