Question

Решить 100б см во вложении

Answer 1

1.

$\sqrt{x}dy=6y^2dx \\ \\ \frac{dy}{y^2}=6 \cdot \frac{dx}{\sqrt{x}} \\ \\ y^{-2} \, dy=6 \cdot x^{-\frac{1}{2}} \, dx \\ \\ \int {y^{-2}} \, dy =6 \int {x^{-\frac{1}{2}}} \, dx \\ \\ -y^{-1}} = 6\cdot \frac{x^\frac{1}{2}}{\frac{1}{2}} +C \\ \\ \frac{1}{y}=-12\sqrt{x}+C \\ \\ y=\frac{1}{C-12\sqrt{x}}$

2.

$\frac{y'\cdot x^3}{y^2}=6 \\ \\ \frac{dy}{dx}\cdot \frac{x^3}{y^2}=6 \\ \\ \frac{dy}{y^2}=\frac{6 \, dx}{x^3} \\ \\ y^{-2} \, dy =6x^{-3} \, dx \\ \\ \int {y^{-2}} \, dy =6 \int {x^{-3}} \, dx \\ \\ -\frac{1}{y}=6\cdot (-\frac{1}{2x^2}) +C \\ \\ \frac{1}{y}=\frac{3}{x^2}+C \\ \\ \frac{1}{y}=\frac{3+Cx^2}{x^2} \\ \\ y=\frac{x^2}{3+Cx^2}$

3.

$y'\cdot \sin^2{x}=y^{-4} \\ \\ \frac{dy}{dx}\cdot \sin^2{x} =\frac{1}{y^{4}} \\ \\ y^4 \, dy =\frac{dx}{\sin^2{x} } \\ \\ \int {y^4} \, dy = \int {\frac{dx}{\sin^2{x}}} \\ \\ \frac{y^5}{5}=-ctg \, x +C \\ \\ y^5=-5 \, ctg \, x +C \\ \\ y=\sqrt[5]{-5 \, ctg \, x +C}$

4.

$y'=2\cos{4x}-5 \\ \\ \frac{dy}{dx}=2\cos{4x}-5 \\ \\ dy=(2\cos{4x}-5)\, dx \\ \\ \int {1} \, dy=\int {(2\cos{4x}-5})} \, dx \\ \\ y=2\cdot \frac{1}{4}\sin{4x}-5x+C \\ \\ y=\frac{1}{2}\sin{4x}-5x+C$

5.

$y'=7x+\cos{3x} \\ \\ \frac{dy}{dx}=7x+\cos{3x} \\ \\ dy = (7x+\cos{3x}) \, dx \\ \\ \int {1} \, dy = \int {(7x+\cos{3x})} \, dx \\ \\ y=7\cdot \frac{x^2}{2}+\frac{1}{3}\sin{3x} +C \\ \\ y=\frac{7}{2}x^2+\frac{1}{3}\sin{3x}+C$