Differential Equations/Substitution 4

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1)

$y' = c s c (x + y) - 1$

$v = x + y$

$v' = 1 + y'$

$v' - 1 = c s c (v) - 1$

$s i n (v) d v = d x$

$\int sin(v)dv=\int dx$

$- c o s (v) = x + C$

$- c o s (x + y) = x + C$

$y = a r c c o s ( - x + C) - x$

2)

$y'=csc(\frac{y}{x})+\frac{y}{x}$

$v=\frac{y}{x}$

$v' x + v = y'$

$v' x + v = c s c (v) + v$

$v' x = c s c (v)$

$sin(v)dv=\frac{dx}{x}$

$\int sin(v)dv=\int \frac{dx}{x}$

$- c o s (v) = l n (x) + C$

$-cos(\frac{y}{x})=ln(x)+C$

$y = x a r c c o s ( - l n (x) + C)$

3)

$y c o s (y 2) y' - s i n (y 2) = 0$

$v = s i n (y 2)$

$v' = 2 y y' c o s (y 2)$

$\frac{v'}{2}-v=0$

$v' = 2 v$

$\frac{dv}{v}=2dx$

$\int \frac{dv}{v}=\int 2dx$

$l n (v) = 2 x + C$

$v = C e 2 x$

$s i n (y 2) = C e 2 x$

$y 2 = a r c s i n (C e 2 x)$

4)

$y' = y l n (y) + y$

$v = l n (y)$

$v'=\frac{y'}{y}$

$v' y = y v + y$

$v' = v + 1$

$\frac{dv}{v+1}=dx$

$\int \frac{dv}{v+1}=\int dx$

$l n (v + 1) = x + C$

$v + 1 = C e x$

$v = C e x - 1$

$l n (y) = C e x - 1$

$y=e^{Ce^x-1}$

5)

$y' = (x 2 + y - 1) 2 - 2 x$

$v = x 2 + y - 1$

$v' = 2 x + y'$

$2 x + y' = (x 2 + y - 1) 2$

$v' = v 2$

$\frac{dv}{v^2}=dx$

$\int \frac{dv}{v^2}=\int dx$

$-\frac{1}{v}=x+C$

$v=\frac{-1}{x+C}$

$x^2+y-1=\frac{-1}{x+C}$

$y=\frac{-1}{x+C}-x^2+1$

6)

$y'=\frac{x^2}{y^2}+\frac{y}{x}$

$v=\frac{y}{x}$

$y' = v + x v'$

$v+xv'=\frac{1}{v^2}+v$

$v^2dv=\frac{dx}{x}$

$\int v^2dv=\int \frac{dx}{x}$

$\frac{1}{3}v^3=ln(x)+C$

$\frac{y^3}{3x^3}=ln(x)+C$

$y=(3x^3(ln(x)+C))^{\frac{1}{3}}$

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Category: Differential Equations

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