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NCERT Exemplar · Q27

Q.Solve: dydx=cos⁡(x+y)+sin⁡(x+y)\frac{dy}{dx}=\cos(x+y)+\sin(x+y). [Hint: Substitute x+y=zx+y=z.]

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The substitution z=x+yz = x+y makes the equation separable. The general solution is tan⁡ ⁣x+y2=Cex−1\tan\!\dfrac{x+y}{2} = C e^{x} - 1.

Substitute z=x+yz = x+y, so dydx=dzdx−1\dfrac{dy}{dx} = \dfrac{dz}{dx} - 1:

dzdx−1=cos⁡z+sin⁡z⇒dzdx=1+cos⁡z+sin⁡z.\frac{dz}{dx} - 1 = \cos z + \sin z \quad\Rightarrow\quad \frac{dz}{dx} = 1 + \cos z + \sin z.

Separate variables:

dz1+cos⁡z+sin⁡z=dx.\frac{dz}{1 + \cos z + \sin z} = dx.

Simplify with half-angles. Using 1+cos⁡z=2cos⁡2z21+\cos z = 2\cos^2\tfrac{z}{2} and sin⁡z=2sin⁡z2cos⁡z2\sin z = 2\sin\tfrac{z}{2}\cos\tfrac{z}{2}:

1+cos⁡z+sin⁡z=2cos⁡z2(cos⁡z2+sin⁡z2),1 + \cos z + \sin z = 2\cos\tfrac{z}{2}\Big(\cos\tfrac{z}{2} + \sin\tfrac{z}{2}\Big),

so

dz1+cos⁡z+sin⁡z=12sec⁡2z21+tan⁡z2 dz.\frac{dz}{1+\cos z+\sin z} = \frac{\tfrac12\sec^2\tfrac{z}{2}}{1 + \tan\tfrac{z}{2}}\,dz.

Integrate. Let u=tan⁡z2u = \tan\tfrac{z}{2}, so du=12sec⁡2z2 dzdu = \tfrac12\sec^2\tfrac{z}{2}\,dz: …

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