Question

Помогите решить пожалуйста))

Answer 1

$1); ; (x^2+x-2)cdot y'=(2x+1)y^3\\frac{dy}{y^3}= frac{(2x+1), dx}{x^2+x-2} \\int y^{-3}, dy=int frac{d(x^2+x-2)}{x^2+x-2}\\frac{y^{-2}}{-2} =ln|x^2+x-2|+C\\ -frac{1}{2y^2}=ln|x^2+x-2|+C\\y^2=frac{1}{C-2cdot ln|x^2+x-2|} \\y=pm frac{1}{sqrt{C2cdot ln|x^2+x-2|}}$

$2); ; (e^{3x}+26)dy=ye^{3x}, dx; ; ,; ; y(0)=26\\ frac{dy}{y}=frac{e^{3x}, dx}{e^{3x}+26}\\int frac{dy}{y}= frac{1}{3}cdot int frac{d(e^{3x}+26)}{e^{3x}+26} \\ln|y|= frac{1}{3}cdot ln|e^{3x}+26|+ln|C|\\y=Ccdot sqrt[3]{e^{3x}+26}\\y(0)=Ccdot sqrt[3]{e^0+26} =26\\Ccdot sqrt[3]{27}=26; ; ,; ; Ccdot 3=26; ,; ; C= frac{26}{3}\\y= frac{26}{3} cdot sqrt[3]{e^{3x}+26}\\P.S. ; ; Esli; ; y(0)=6,; ; to; ; C=frac{6}{3}=2; ; to ; ; y=2cdot sqrt[3]{e{3x}+26}$

$3); ; y'cdot 2^{x}cdot sin^3y=x\\ frac{dy}{dx}cdot sin^3y = frac{x}{2^{x}} \\int sin^3y, dy=int 2^{-x}, x, dx\\int sin^3y, dy=int sin^2ycdot siny, dy=-int (1-cos^2y)cdot d(cosy)=\\=-int d(cosy)+int cos^2y, d(cosy)=-cosy+frac{cos^3y}{3}+C_1\\int 2^{-x}, x, dx=[, u=x,; du=dx,; dv=2^{-x}dx,; v=-frac{2^{-x}}{ln2}, ]=\\=-frac{xcdot 2^{-x}}{ln2}+frac{1}{ln2}int 2^{-x}, dx=- frac{xcdot 2^{-x}}{ln2}-frac{2^{-x}}{ln^22} +C_2\\\-cosy+ frac{cos^3y}{3}=-frac{2^{-x}}{ln^22}cdot (1+xcdot ln2)+C^{*}; ,; ; C^{*}=C_2-C_1$

$(-3cosy+cos^3y)cdot ln^22=-3cdot 2^{-x}(1+xcdot ln2)+C; ; ,; ; C=3, ln^22cdot C^{*}$