Question

Pls help me with this question.the question is in the attachment

Answer 1

Step-by-step explanation:

We have,

$\tt{ \dfrac{dy}{dx} = cos(x + y) }$

$\bf{Put \: \: \: \: x + y = v}$

$\bf{ \implies 1 + \dfrac{dy}{dx} = \dfrac{dv}{dx}}$

So,

$\tt{ \dfrac{dv}{dx} - 1= cos(v) }$

$\tt{ \implies\dfrac{dv}{dx} = cos(v) + 1 }$

$\tt{ \implies\dfrac{dv}{1 + cos(v) } = dx }$

Integrating both sides,

$\displaystyle \tt{ \implies \int\dfrac{dv}{1 + cos(v) } = \int dx }$

$\displaystyle \tt{ \implies \int\dfrac{ \left( 1 - cos(v) \right)dv}{ \left(1 + cos(v) \right) \left(1 - cos(v) \right)} = x + c_{1}}$

$\displaystyle \tt{ \implies \int\dfrac{ \left( 1 - cos(v) \right)dv}{ 1 - cos^{2} (v) } = x + c_{1}}$

$\displaystyle \tt{ \implies \int\dfrac{ 1 - cos(v) }{ sin^{2} (v) } \: dv = x + c_{1}}$

$\displaystyle \tt{ \implies \int\dfrac{ 1 }{ sin^{2} (v) } \: dv - \int\dfrac{ cos(v) }{ sin^{2} (v) } \: dv = x + c_{1}}$

$\displaystyle \tt{ \implies \int cosec^{2} (v) \: dv - \int \: cosec(v) \: cot(v) \: dv = x + c_{1}}$

$\displaystyle \tt{ \implies - cot (v) +cosec(v) +c_{2} = x + c_{1}}$

$\displaystyle \tt{ \implies - cot (x + y) +cosec(x + y) = x + k}$

Answer 2

Given Question :-

Solve the differential equation

$\purple{\rm :\longmapsto\:\dfrac{dy}{dx} = cos(x + y)}$

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

Given differential equation is

$\purple{\rm :\longmapsto\:\dfrac{dy}{dx} = cos(x + y)}$

To solve this Differential equation, we use the Method of Substitution.

So, Substitute

$\rm :\longmapsto\:x + y = z$

$\rm :\longmapsto\:\dfrac{d}{dx}(x + y) = \dfrac{d}{dx}z$

$\rm :\longmapsto\:1 + \dfrac{dy}{dx} = \dfrac{dz}{dx}$

$\purple{\rm :\longmapsto\: \dfrac{dy}{dx} = \dfrac{dz}{dx} - 1}$

So, on substituting these values, the differential equation can be rewritten as

$\rm :\longmapsto\:\dfrac{dz}{dx} - 1 = cosz$

$\rm :\longmapsto\:\dfrac{dz}{dx} =1 + cosz$

$\rm :\longmapsto\:\dfrac{dz}{1 + cosz} = dx$

can be further rewritten as

$\rm :\longmapsto\:\dfrac{dz}{2 {cos}^{2} \dfrac{z}{2} } = dx$

$\rm :\longmapsto\:\dfrac{1}{2} {sec}^{2}\dfrac{z}{2}dz = dx$

On integrating both sides, we get

$\rm :\longmapsto\:\dfrac{1}{2} \displaystyle\int {sec}^{2}\dfrac{z}{2}dz = \displaystyle\int \: dx$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int \: {sec}^{2}(ax + b) \: dx = \frac{tan(ax + b)}{a} + c}}}$

So, using this, we get

$\rm :\longmapsto\: tan \dfrac{z}{2} = x + c$

On substituting the value of z, we get

$\rm :\longmapsto\: tan \bigg(\dfrac{x + y}{2} \bigg)= x + c$

Hence, Option (3) is correct.

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LEARN MORE

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