Question

If x^3dy+xydx=2ydx+x^2dy and y(2) = e then y(4) =? ​

Answer 1

$\bf \huge \underline \red{Answer : - }$

$\sf \: {x}^{3} dy + xydx = {x}^{2} dy + 2ydx \\ \\ \sf \: {x}^{2} (x - 1)dy = y(2 - x)dx \\ \\\sf \: \int \: \frac{dy}{y} = \int \: \frac{(2 - x)}{ {x}^{2}(x - 1) } dx \\ \\ \sf \:ln \: y = \int \: \left( \dfrac{a}{x} + \frac{b}{ {x}^{2} } + \frac{c}{x - 1} \right) \\ \\ \sf \:a = - 1 \: \: \: \: and \: \: b = - 2 \: \: and \: \: c = 1 \\ \\ \sf \:ln \: y = - 1ln \: x + \frac{2}{x} + ln(x - 1) + c \\ \\ \sf \:ln \: y = ln \: ( \frac{x - 1}{x} ) + ( \frac{2}{x} ) + c \\ \\ \sf \:y(2) = e \\ \\ \sf \:x = 2 \: \: and \: \: \: y = e \\ \\ \sf \: \cancel {1} = - ln \: 2 \: \: \cancel{ + 1} + c \\ \\ \sf \:c = ln \: 2 \\ \\ \sf \: ln \: y = ln \: ( \frac{x - 1}{x} ) + \frac{2}{n} ln \: 2 \\ \\ \sf \:ln \: y = ln \: ( \frac{3}{4} ) + 2 + ln \: 2 \\ \\ \sf \implies \:ln \: ( \frac{11}{4} ) + 2 \\ \\ \sf \: ln \: y = ln \: ( \frac{ \frac{11}{4} }{ {e}^{2} } ) \\ \\\sf \implies \: y = \frac{11}{ {4e}^{2} }$

Answer 2

GIVEN :–

• Differentiate equation –

$\\\bf \implies {x}^{3} dy + xydx = 2ydx + {x}^{2} dy$

$\\\bf \implies y(2) = e$

TO FIND :–

$\\\bf \implies y(4) =?$

SOLUTION :–

$\\\bf \implies {x}^{3} dy + xydx = 2ydx + {x}^{2} dy$

$\\\bf \implies {x}^{3} dy - {x}^{2}dy = 2ydx - xydx$

$\\\bf \implies ({x}^{3} - {x}^{2})dy = y(2 - x)dx$

$\\\bf \implies \dfrac{dy}{y} = \dfrac{(2 - x)}{( {x}^{3} - {x}^{2}) }dx$

• Now Integrate on both sides –

$\\\bf \implies \int \dfrac{dy}{y} = \int \dfrac{(2 - x)}{( {x}^{3} - {x}^{2}) }.dx$

$\\\bf \implies \int \dfrac{dy}{y} = \int \dfrac{(2 - x)}{ {x}^{2} (x-1)}.dx$

• By using Partial fraction method , We should write this as –

$\\\bf \implies \dfrac{(2 - x)}{ {x}^{2} (x-1)} = \dfrac{A}{x} +\dfrac{B}{ {x}^{2} } + \dfrac{C}{(x - 1)}$

$\\\bf \implies \dfrac{(2 - x)}{ {x}^{2} (x-1)} = \dfrac{Ax}{ {x}^{2} } +\dfrac{B}{ {x}^{2} } + \dfrac{C}{(x - 1)}$

$\\\bf \implies \dfrac{(2 - x)}{ {x}^{2} (x-1)} = \dfrac{Ax +B }{ {x}^{2} } + \dfrac{C}{(x - 1)}$

$\\\bf \implies \dfrac{(2 - x)}{ {x}^{2} (x-1)} = \dfrac{(Ax +B)(x - 1) +C( {x}^{2}) }{ {x}^{2}(x - 1)}$

$\\\bf \implies (2 - x)=(Ax +B)(x - 1) +C( {x}^{2})$

$\\\bf \implies (2 - x)=A {x}^{2} - Ax + Bx -B+C( {x}^{2})$

$\\\bf \implies (2 - x)=(A + C) {x}^{2} +(B- A)x -B$

• After comparing –

$\\\bf \implies A = - 1, B = - 2\: and \: C = 1$

• So that –

$\\\bf \implies \dfrac{(2 - x)}{ {x}^{2} (x-1)} = \dfrac{ - 1}{x} +\dfrac{ - 2}{ {x}^{2} } + \dfrac{1}{(x - 1)}$

• And integration –

$\\\bf \implies \int \dfrac{dy}{y} = \int \dfrac{ - 1}{x}.dx+ \int\dfrac{ - 2}{ {x}^{2} } .dx+ \int \dfrac{1}{(x - 1)}.dx$

$\\\bf \implies ln(y) = - ln(x)+ \dfrac{2}{ {x}} +ln(x - 1) + c$

• When x = 2 & y = e :–

$\\\bf \implies ln(e) = - ln(2)+ \dfrac{2}{2} +ln(2- 1) + c$

$\\\bf \implies 1 = - ln(2)+1+ln(1) + c$

$\\\bf \implies \large { \boxed{ \bf c= ln(2)}}$

• Now when x = 4 –

$\\\bf \implies ln(y) = - ln(4)+ \dfrac{2}{4} +ln(4- 1) + ln(2)$

$\\\bf \implies ln(y) = - ln(2^2)+ \dfrac{1}{2} +ln(3) + ln(2)$

$\\\bf \implies ln(y) = -2 ln(2)+ \dfrac{1}{2} +ln(3) + ln(2)$

$\\\bf \implies ln(y) = \dfrac{1}{2} +ln(3) - ln(2)$

$\\\bf \implies ln(y) = ln(\sqrt{e}) +ln \left(\dfrac{3}{2}\right)$

$\\\bf \implies ln(y) = ln\left[(\sqrt{e}) \left(\dfrac{3}{2}\right)\right]$

$\\\large\implies{\boxed{\bf y= \dfrac{3\sqrt{e}}{2}}}$