22. Solve (e^x+e^{-x})dy/dx=y^2 be separable method

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22. Solve (e^x+e^{-x})dy/dx=y^2 be separable method

To solve the differential equation \left(e^x+e^{-x}\right)\frac{dy}{dx}=y^2 by separable method, we need to rearrange it so that all the y terms are on one side and all the x terms are on the other side. Then, we can integrate both sides to obtain the solution.

First, we can multiply both sides by dx and divide both sides by y^2 to get:

    \[\frac{dy}{y^2} = \frac{dx}{e^x+e^{-x}}\]

Now, we can integrate both sides with respect to their respective variables:

    \[\int \frac{dy}{y^2} = \int \frac{dx}{e^x+e^{-x}}\]

To integrate the left-hand side, we can use the power rule:

    \[\int \frac{dy}{y^2}=-\frac{1}{y}+{C_1}\]

where C_1 is constant of integration.

To integrate the right-hand side, we can use the substitution u=e^x and du=e^xdx, which gives us:

    \[\int \frac{dx}{e^x+e^{-x}} = \int \frac{du/u}{u+1/u} = \int \frac{du}{u^2+1} = \tan^{-1}(u) + C_2\]

where C_2 is another constant of integration.

Substituting u=e^x back into the equation and simplifying, we get:

    \[-\frac{1}{y} +C_1 = \tan^{-1}(e^x) + C_2\]

    \[-\frac{1}{y} = \tan^{-1}(e^x) + C\]

Multiplying both sides by y, we get:

    \[y\left(\tan^{-1}(e^x) +C\right) = -1\]

Therefore, the general solution to the differential equation is:

    \[y\left(\tan^{-1}(e^x) +C\right) = -1\]

    \[y=-\frac{1}{\left(\tan^{-1}(e^x) +C\right)}\]

where C is an arbitrary constant of integration.

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