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Exercise 9.5 · Q8

Q.Solve the following differential equation: (1+x2)dy+2xy dx=cot⁡x dx(x≠0)(1 + x^2) dy + 2xy \ dx = \cot x \ dx \quad (x \ne 0)

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Linear first-order ODE; the left side is an exact derivative. Solution: y=log⁡∣sin⁡x∣+C1+x2y = \dfrac{\log|\sin x| + C}{1 + x^2}.

Write the equation with dydx\dfrac{dy}{dx}:

(1+x2)dydx+2xy=cot⁡x.(1 + x^2)\frac{dy}{dx} + 2xy = \cot x.

The left side is exactly the derivative of a product:

ddx[(1+x2) y]=(1+x2)dydx+2xy.\frac{d}{dx}\big[(1 + x^2)\,y\big] = (1 + x^2)\frac{dy}{dx} + 2xy.

So the equation becomes

ddx[(1+x2) y]=cot⁡x.\frac{d}{dx}\big[(1 + x^2)\,y\big] = \cot x.

Integrate both sides with respect to xx: …

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