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Question 150 of 177

Q.Solve: dydx=cos⁡(x+y)\dfrac{dy}{dx} = \cos(x + y)

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2017Subjective· 3mImportance★★★★★
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Substitute v=x+yv=x+y to reduce to a separable equation.

dydx=cos⁡(x+y)\dfrac{dy}{dx} = \cos(x+y)

Let v=x+y⇒dvdx=1+dydx⇒dydx=dvdx−1v = x+y \Rightarrow \dfrac{dv}{dx} = 1+\dfrac{dy}{dx} \Rightarrow \dfrac{dy}{dx} = \dfrac{dv}{dx}-1

Substituting:

dvdx−1=cos⁡v  ⟹  dvdx=1+cos⁡v=2cos⁡2(v2)\dfrac{dv}{dx}-1 = \cos v \implies \dfrac{dv}{dx} = 1+\cos v = 2\cos^2\left(\dfrac v2\right)

Separating variables:

dv2cos⁡2(v/2)=dx  ⟹  12sec⁡2(v2)dv=dx\dfrac{dv}{2\cos^2(v/2)} = dx \implies \dfrac12\sec^2\left(\dfrac v2\right)dv = dx

Integrating both sides: …

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