Question

Solve dy/dx + ycotx = cosx​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Differential equation is

$\rm \: \dfrac{dy}{dx} + y \: cotx \: = \: cosx$

Its a linear differential equation, so on comparing with

$\rm \: \dfrac{dy}{dx} + py \: = \: q, \: we \: get$

$\rm \: p \: = \: cotx$

$\rm \: q \: = \: cosx$

Now, Integrating Factor is

$\rm \: I.F. \: = \: {e}^{\displaystyle \int \rm pdx}$

$\rm \: I.F. \: = \: {e}^{\displaystyle \int \rm cotx \: dx}$

$\rm \: I.F. \: = \: {e}^{log \: sinx}$

$\rm \: I.F. \: = \: sinx$

So, Solution of differential equation is given by

$\rm \: y \times I.F. \: = \: \displaystyle \int \rm q \times I.F. \: dx$

$\rm \: y \times sinx \: = \: \displaystyle \int \rm cosx \times sinx \: dx$

$\rm \: y \: sinx \: = \: \displaystyle \int \rm sinx \: cosx \: dx$

$\rm \: y \: sinx \: = \: \dfrac{1}{2} \displaystyle \int \rm \: 2 \: sinx \: cosx \: dx$

$\rm \: y \: sinx \: = \: \dfrac{1}{2} \displaystyle \int \rm \: sin2x \: dx$

$\rm \: y \: sinx \: = \: - \dfrac{1}{2} \times \dfrac{cos2x}{2} + c$

$\rm \: y \: sinx \: = \: - \dfrac{cos2x}{4} + c$

$\rm \: 4y \: sinx \: = \: - cos2x + 4c$

$\rm \: 4y \: sinx \: = \: - cos2x + c' \: \: \: \: \: \{where \: c' \: = \: 4c \}$

Hence,

$\color{green}\rm\implies \:\boxed{ \rm{4ysinx= -cos2x + c' \: \: \{where \: c' \: = \: 4c \} }}$

$\rule{190pt}{2pt}$

Additional Information :-

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

Answer 2

Answer:

4ysinx=-cos2x+4c

Step-by-step explanation:

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