Category Archives: Math

Math

Solve Differential Equation, $ \frac{{dy}}{{dx}} = \sqrt {x + y} $

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Let, $ \sqrt {x + y} = t$

So, $ \frac{1}{{2\sqrt {x + y} }}\left( {1 + \frac{{dy}}{{dx}}} \right) = \frac{{dt}}{{dx}}$

Or, $\frac{1}{{2t}}\left( {1 + \frac{{dy}}{{dx}}} \right) = \frac{{dt}}{{dx}}$

$\therefore \frac{{dy}}{{dx}} = 2t\frac{{dt}}{{dx}} – 1$

From the original differential equation,

$2t\frac{{dt}}{{dx}} – 1 = t$

$\therefore 2t\frac{{dt}}{{dx}} = 1 + t$

Or, $\frac{{2t}}{{1 + t}}dt = dx$

$ \Rightarrow 2\int {\frac{{1 + t – 1}}{{1 + t}}dt} = \int {dx} $

$ \Rightarrow 2\int {\left( {1 – \frac{1}{{1 + t}}} \right)dt} = \int {dx} $

$ \Rightarrow 2t – 2\ln (1 + t) = x + C$

$ \Rightarrow 2\sqrt {x + y} – 2\ln (1 + \sqrt {x + y} ) = x + C$

Or, $\sqrt {x + y} – \ln (1 + \sqrt {x + y} ) = \frac{x}{2} + C’$

Math

Given $x+y=2$ & $xy=3$, evaluate $x^5+y^5$

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Question

Given x+y=2 & xy=3, evaluate $x^5+y^5$.

Solution

Putting y=2-x in the equation xy=3,

x(2-x)=3

Or, x2-2x+3=0

Or, $x=\frac {2\pm\sqrt {4-12}}{2}$

Or, $x=1\pm i.\sqrt 2$

Out of these two complex conjugate roots, one should represent x and the other should represent y.

Let $x=1+ i.\sqrt 2$, So $y=1- i.\sqrt 2$

${x^5} + {y^5} = {(1 + i\sqrt 2 )^5} + {(1 – i\sqrt 2 )^5}$

$ = [{}^5{C_0} + {}^5{C_1}i\sqrt 2 + {}^5{C_2}{(i\sqrt 2 )^2} + {}^5{C_3}{(i\sqrt 2 )^3} + {}^5{C_4}{(i\sqrt 2 )^4} + {}^5{C_5}{(i\sqrt 2 )^5}]$$+$$[{}^5{C_0} – {}^5{C_1}i\sqrt 2 + {}^5{C_2}{(i\sqrt 2 )^2} – {}^5{C_3}{(i\sqrt 2 )^3} + {}^5{C_4}{(i\sqrt 2 )^4} – {}^5{C_5}{(i\sqrt 2 )^5}]$

$= 2[{}^5{C_0} + {}^5{C_2}{(i\sqrt 2 )^2} + {}^5{C_4}{(i\sqrt 2 )^4}]$

$ = 2\left[ {1 + \frac{{5!}}{{2!3!}}( – 1)(2) + 5(4)} \right]$

=2(1-20+20)

=2