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Popular >

(dy)/(dx)y=arccos(1/x)

Solution

\frac{d y}{d x} y = arccos (\frac{1}{x})

Solution

y = 2 x arccos (\frac{1}{x}) - ln (\frac{x ^{2} - 1}{x} + 1) + ln (\frac{x ^{2} - 1}{x} - 1) + c_{1}, y = - 2 x arccos (\frac{1}{x}) - ln (\frac{x ^{2} - 1}{x} + 1) + ln (\frac{x ^{2} - 1}{x} - 1) + c_{1}

Solution steps

\frac{d y}{d x} y = arccos (\frac{1}{x})

Solve separable ODE:

y = 2 x arccos (\frac{1}{x}) - ln (\frac{x ^{2} - 1}{x} + 1) + ln (\frac{x ^{2} - 1}{x} - 1) + c_{1}, y = - 2 x arccos (\frac{1}{x}) - ln (\frac{x ^{2} - 1}{x} + 1) + ln (\frac{x ^{2} - 1}{x} - 1) + c_{1}

y = 2 x arccos (\frac{1}{x}) - ln (\frac{x ^{2} - 1}{x} + 1) + ln (\frac{x ^{2} - 1}{x} - 1) + c_{1}, y = - 2 x arccos (\frac{1}{x}) - ln (\frac{x ^{2} - 1}{x} + 1) + ln (\frac{x ^{2} - 1}{x} - 1) + c_{1}

Popular Examples

y^{^{'}} - \frac{7}{5} y = 0

y^'+y=y(te^{t^2}+1)t,y(0)=1

y^{^{'}} + y = y (t e^{t^{2}} + 1) t, y (0) = 1

(dy)/(dx)y=(x^{x-3})

\frac{d y}{d x} y = (x^{x - 3})

((x^3+1)/(x^2))(dy)/(dx)= 1/y ,y(0)=-1

(\frac{x ^{3} + 1}{x ^{2}}) \frac{d y}{d x} = \frac{1}{y}, y (0) = - 1

e^{x+y}*(dy)/(dx)=x

e^{x + y} \cdot \frac{d y}{d x} = x

Frequently Asked Questions (FAQ)

What is the solution for (dy)/(dx)y=arccos(1/x) ?
The solution for (dy)/(dx)y=arccos(1/x) is