Question

Помогите пожалуйста!!!

Answer 1

$1)\ \ y^2\, dy=e^{x}\, dx\ \ ,\ \ y(0)=1\\\\\int y^2\, dy=\int e^{x}\, dx\ \ ,\ \ \ \dfrac{y^3}{3}=e^{x}+C\ \ ,\\\\y(0)=1:\ \ \dfrac{1^3}{3}=\underbrace {e^{0}}_{1}+C\ \ ,\ \ C=\dfrac{1}{3}-1=-\dfrac{2}{3}\\\\\dfrac{y^3}{3}=e^{x}-\dfrac{2}{3}\ \ ,\ \ \ \underline {y^3=3e^{x}-2\ }$

$2)\ \ A\cdot B=\left(\begin{array}{ccc}7&2\\2&4\end{array}\right)\cdot \left(\begin{array}{ccc}1&-2\\6&3\end{array}\right)=\left(\begin{array}{ccc}19&-8\\26&8\end{array}\right)\\\\\\C_{11}=7\cdot 1+2\cdot 6=19\ \ ,\ \ C_{12}=7\cdot (-2)+2\cdot 3=-8\\\\C_{21}=2\cdot 1+4\cdot 6=26\ \ ,\ \ C_{22}=2\cdot (-2)+4\cdot 3=8$

$3)\ \ i^2=-1\\\\Z=\dfrac{5i}{1+3i}+(-1+2i)^2=\dfrac{5\, i\, (1-3i)}{1-(3i)^2}+1-4i+4i^2=\\\\\\=\dfrac{5i-15i^2}{1+9}+1-4i-4=\dfrac{5i+15}{10}-3-4i=\dfrac{3}{2}+\dfrac{1}{2}\, i-3-4i=\\\\\\=-\dfrac{3}{2}+\dfrac{7}{2}\, i=-1,5+3,5\, i$