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- Thread starter fullonoob
- Start date
dy/dx=((log(x)-1)/(log(x))^2)Show that y = x/logx is a solution of the eqation dy/dx = (y/x) - (y/x)^2
Thx
=1/log(x)-1/(log(x))^2
y=x/log(x)
y/x=1/log(x)
Thus, dy/dx=(y/x)-(y/x)^2
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y = x/log xShow that y = x/logx is a solution of the eqation dy/dx = (y/x) - (y/x)^2
Show that y = log(logx) is a solution of the equation x d^2y/dx^2 + x(dy/dx)^2 + dy/dx = 0
Thx
y/x = 1/log x
dy/dx = (log x - 1)/(log x)²
LHS = dy/dx
= (log x - 1)/(log x)²
= log x/(log x)² - 1/(log x)²
=1/log x - (1/log x)²
= (y/x) - (y/x)²
= RHS
When y = log(log x)
dy/dx = 1/(x log x)
d²y/dx² = - 1/(x² log x) - (1/x²(log x)²)
= - (log x + 1)/(x²(log x)²)
LHS = x(d²y/dx²) + x(dy/dx)² + dy/dx
= - (log x + 1)/(x(log x)²) + 1/(x(log x)²) + 1/(x log x)
= - log x /(x(log x)²) + 1/(x log x)
= - 1/(x log x) + 1/(x log x)
= 0
= RHS
=(1/log(x)) - (1/(log(x))^2)what happened there?
=(log(x)-1) changed to 1/ log(x) -1
lol ty a few minor errors there but its ok\sqrt{x-1}}{x+2} \\ ln(y)= ln(\frac{(x+1)\sqrt{x-1}}{x+2})\\ ln(y) =ln(x+1)+\frac{1}{2}ln(x-1)-ln(x-1)\\ \frac{1}{y}\frac{dy}{dx}=\frac{1}{x+1}+\frac{1}{2(x-1)}-\frac{1}{x-1} \\ \text{multiplying both sides by y, canceling common factors} \\ \frac{dy}{dx}= \frac{\sqrt{x-1}}{x+2}+\frac{x+1}{2(x+2)\sqrt{x-1}}-\frac{(x+1)\sqrt{x-1}}{(x+2)^2})
Really?, damn. D: i guess im abit tired. probably shouldn;t write an english essay at the same time as doing math then. ><lol ty a few minor errors there but its ok
edit: wait nvm, i saw it.
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