Question

y′+(cotx)y=xcscx,y(^/2)=1
o y=−x^2/2sinx−1−π^2/8 sinx
y=x^2/2sinx+1−π^2/8/sinx
y=−x^2/2sinx+1+π^2/8/sinx
y=−x^2/sinx−1−π^2/8/2sinx

Answer 1

Answer on Question #78899 – Math – Differential Equations

Question

$$y' + (\cot x)y = x \csc x, \quad y\left(\frac{\pi}{2}\right) = 1$$ $$y = -\frac{x^2}{2 \sin x} - \frac{1 - \frac{\pi^2}{8}}{\sin x}$$ $$y = \frac{x^2}{2 \sin x} + \frac{1 - \frac{\pi^2}{8}}{\sin x}$$ $$y = -\frac{x^2}{2 \sin x} + \frac{1 + \frac{\pi^2}{8}}{\sin x}$$ $$y = -\frac{x^2}{2 \sin x} - \frac{1 - \frac{\pi^2}{8}}{\sin x}$$

Solution

We have a linear first order differential equation

$$\frac{dy}{dx} + (\cot x)y = x \csc x$$ $$P(x) = \cot x, \quad Q(x) = x \csc x$$

Integrating factor

$$IF = e^{\int P(x)dx} = e^{\int \cot x\, dx}$$

Indefinite integral

$$\int \cot x\, dx = \int \frac{\cos x}{\sin x}\, dx$$

Substitution

$$u = \sin x, \quad du = \cos x\, dx$$ $$\int \cot x\, dx = \int \frac{\cos x}{\sin x}\, dx = \int \frac{1}{u}\, du = \ln |u| + C_1 = \ln |\sin x| + C_1$$ $$IF = e^{\int P(x)dx} = e^{\int \cot x\, dx} = e^{\ln (\sin x)} = \sin x$$ $$\sin x \frac{dy}{dx} + \sin x (\cot x)y = x \sin x \csc x$$ $$\frac{d}{dx} (y \sin x) = x \sin x \csc x$$ $$\frac{d}{dx} (y \sin x) = x$$

Integrate

$$y \sin x = \frac{x^2}{2} + C$$ $$y \left(\frac {\pi}{2}\right) = 1 \Rightarrow 1 \left(\sin \frac {\pi}{2}\right) = \frac {\left(\frac {\pi}{2}\right) ^ {2}}{2} + C \Rightarrow C = 1 - \frac {\pi^ {2}}{8}$$ $$y \sin x = \frac {x ^ {2}}{2} + 1 - \frac {\pi^ {2}}{8}$$ $$y = \frac {x ^ {2}}{2 \sin x} + \frac {1 - \frac {\pi^ {2}}{8}}{\sin x}$$

Answer: $y = \frac{x^2}{2\sin x} + \frac{1 - \frac{\pi^2}{8}}{\sin x}$.

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