Question

find dy /dx =5x-2x^2​

Answer 1

Answer:3(1−x)

5/9

5

​

(3−x)−2sin(4x+2)

Given, y=5x(1−x)

−2/3

+cos

2

(2x+1)

dx

dy

​

=5.1(1−x)

−2/3

+5x.(−

32

)(1−x)

−2/3−1

(−1)+2cos(2x+1).{−sin(2x+1)}.2.=

3(1−x)

5/95

(3−x)−2sin(4x+2).

Step-by-step explanation:

Answer 2

Answer:

$\mathrm{y = \dfrac{5x^2}{2} - \dfrac{2x^3}{3} + c}$

Step-by-step explanation:

Given :

$\mathrm{\dfrac{dy}{dx} = 5x - 2x^2}$

To find :

Solution of the given DE

Solution :

$\longmapsto \mathrm{\dfrac{dy}{dx} = 5x - 2x^2}$

$\implies \mathrm{dy = (5x-2x^2)dx}$

Integrating on both sides,

$\implies \mathrm{\int dy = \int (5x-2x^2)dx}$

We know that,

$\boxed{\mathrm{\int x^n dx = \dfrac{x^{n+1}}{n+1} + c}}$

So,

$\implies \mathrm{y = \dfrac{5x^2}{2} - \dfrac{2x^3}{3} + c}$

$\therefore \underline{\boxed{\mathbf{y = \dfrac{5x^2}{2} - \dfrac{2x^3}{3} + c}}}$

❖ Extra information :

$\bigstar \textsf{ \underline{Some Basic Integrals :}}$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx+c \\\\ \sf x^n \ (n \neq -1)& \sf \dfrac{x^{n+1}}{n+1} + c \\\\ \sf \dfrac{1}{x} & \sf logx+ c\\\\ \sf {e}^{x} & \sf {e}^{x}+c\\\\ \sf sinx & \sf - \: cosx+ c \\\\ \sf cosx & \sf \: sinx + c\\\\ \sf {sec}^{2} x & \sf tanx + c\\\\ \sf {cosec}^{2}x & \sf - cotx+ c \\\\ \sf secx \: tanx & \sf secx + c\\\\ \sf cosecx \: cotx& \sf -\: cosecx + c\end{array}} \\ \end{gathered}$

Hope it helps!!