Tolong dibantu ya soalnya senin besok dikumpulin :(

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Tolong dibantu ya soalnya senin besok dikumpulin :(

Tolong dibantu ya soalnya senin besok dikumpulin 🙁

Jawaban:

di bawah

Penjelasan dengan langkah-langkah:

15

 frac{dy}{dx} = 3 + 2y \ frac{dy}{3 + 2y} = dx \ int{ frac{dy}{3 + 2y} }dy = int{}dx \ frac{1}{2} ln(3 + 2y) + c_{1} = x + c_{2} \ frac{1}{2} ln(3 + 2y) = x + c

16.

5 frac{dy}{dx} = cot(2y) \ 5 tan(2y) dy = dx \ 5int{ tan(2y) }dy = int{}dx \ frac{5}{2} int{ tan(2y) }d(2y) = int{}dx \ frac{5}{2} ln( sec(x) ) + c_{1} = x + c_{2} \ frac{5}{2} ln( sec(x) ) = x + c

17.

y frac{dy}{dx} = 3 - {y}^{2} \ frac{y}{3 - {y}^{2} } dy = dx \ int{ frac{y}{3 - {y}^{2} } }dy = int{}dx \ int{ frac{y}{3 - {y}^{2} } } frac{d( 3 - {y}^{2} )}{ - 2y} = int{}dx \ - frac{1}{2} int{ frac{1}{3 - {y}^{2} } }d(3 - {y}^{2} ) = int{}dx \ - frac{1}{2} ln(3 - {y}^{2} ) + c_{1} = x + c_{2} \ - frac{1}{2} ln(3 - {y}^{2} ) = x + c

18.

2 frac{dy}{dx} + 3y = 4 \ 2 frac{dy}{dx} = 4 - 3y \ frac{2}{4 - 3y} dy = dx \ 2int{ frac{1}{4 - 3y} }dy = int{}dx \ 2int{ frac{1}{4 - 3y} } frac{d(4 - 3y)}{ - 3} = int{}dx \ - frac{2}{3} int{ frac{1}{4 - 3y} }d(4 - 3y) = int{}dx \ - frac{2}{3} ln(4 - 3y) + c_{1} = x + c_{2} \ - frac{2}{3} ln(4 - 3y) = x + c

19.

 frac{dy}{dx} = 2 tan(y) \ cot(y) dy = 2dx \ int{ cot(y) }d y = int{2}dx \ ln( sin(y) ) + c_{1} = 2x + c_{2} \ ln( sin(y) ) = 2x + c \ sin(y) = {e}^{2x + c} \ y = arcsin( {e}^{2x + c} )

20.

y frac{dy}{dx} = 1 - y, : y(1) = 0 \ frac{y}{1 - y} dy = dx \ - dy + frac{1}{1 - y} dy = dx \ - int{}dy + int{ frac{1}{1 - y} }dy = int{}dx \ - int{}dy - int{ frac{1}{1 - y} }d(1 - y) = int{}dx \ - y - ln(1 - y) + c_{1} = x + c_{2} \ x + y + ln(1 - y) = c \ y(1) = 0 \ 1 + 0 + ln(1) = c \ c = 1 \ x + y + ln(1 - y) = 1

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