Question

небольшие 2 задания, с обьяснением
27 и 28

Answer 1

$27)\; \; \frac{3\, dy}{y}-\frac{dx}{x}\; \; ,\; \; \; y(1)=\sqrt2\\\\\int \frac{dy}{y}=\int \frac{dx}{x}\; \; ,\; \; ln|y|=ln|x|+ln|C|\; \; ,\; \; y=Cx\\\\y(1)=\sqrt2:\; \; \sqrt2=C\cdot 1\; \; \; \to \; \; \; C=\sqrt2\\\\\underline {\; y=\sqrt2\, x\; }$

$28)\; \; y\cdot y'=sin5x\; \; \; \to \; \; \; \dfrac{dy}{dx}=\dfrac{sin5x}{y}\\\\\\\int y\, dy=\int sin5x\, dx\\\\\dfrac{y^2}{2}=-\dfrac{1}{5}\cdot cos5x+C$

$29)\; \; \dfrac{y\, dy}{dx}=4-x^2\; \; ,\; \; y(0)=1\\\\\\\int y\, dy=\int (4-x^2)\, dx\\\\\dfrac{y^2}{2}=4x-\dfrac{x^3}{3}+C\; \; \; \to \; \; \; y^2=8x-\dfrac{2x^3}{3}+C\\\\y(0)=1:\; \; 1^2=2\cdot 0-\dfrac{2\cdot 0^3}{3}+C\; \; ,\; \; C=1\; \; \; \Rightarrow \\\\y^2=8x-\dfrac{2x^3}{3}+1\; \; \; ili\; \; \; \; y=\pm \sqrt{8x-\dfrac{2x^3}{3}+1}$