у' = 2 cos(x/2+[tex] \pi [/tex]/1)у' = 4/(3x-1)^2. Created by shmatov1994. algebra-ru

(3x-1)^2...

shmatov1994 @shmatov1994

August 2022 1 49 Report

у' = 2 cos(x/2+[tex] \pi [/tex]/1)
у' = 4/(3x-1)^2

Answers & Comments

Nik133 $y=\int{2cos(\frac x2+\pi)}dx=-2\int{cos\frac x2}dx=-2*2sin\frac x2+C=-4sin\frac x2+C \\ \\ \\
y=\int{\frac{4}{(3x-1)^2}}dx=4*\frac13\int{(3x-1)^{-2}}d(3x-1)=\frac43*\frac{(3x-1)^{-2+1}}{-2+1}+C= \\ \\ =-\frac{4}{3(3x-1)}+C=\frac{4}{1-9x}+C$

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