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Question 1

Question 2

Решите задачу:

$1)\; \; (x^2+x-2)\cdot y'=(2x+1)y^3\\\\\frac{dy}{y^3}= \frac{(2x+1)\, dx}{x^2+x-2} \\\\\int y^{-3}\, dy=\int \frac{d(x^2+x-2)}{x^2+x-2}\\\\\frac{y^{-2}}{-2} =ln|x^2+x-2|+C\\\\ -\frac{1}{2y^2}=ln|x^2+x-2|+C\\\\y^2=\frac{1}{C-2\cdot ln|x^2+x-2|} \\\\y=\pm \frac{1}{\sqrt{C2\cdot ln|x^2+x-2|}}$

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

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

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