Mathematics - Differential Equation Question with Solution | TestHub

The given differential equation is a Bernoulli equation:

$$\frac{dy}{dx}-y\tan x=-y^2\sec x$$

Divide the entire equation by $y^2$ to prepare for substitution:

$$\frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}\tan x=-\sec x$$

Introduce a new variable $v=\frac{1}{y}$. Then,

$$\frac{dv}{dx}=-\frac{1}{y^2}\frac{dy}{dx}$$

Substituting these into the equation yields a first-order linear differential equation:

$$-\frac{dv}{dx}-v\tan x=-\sec x$$

$$\frac{dv}{dx}+v\tan x=\sec x$$

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Step 2: Solve the Linear Equation

We solve the linear ODE using an integrating factor (IF), $I(x)$:

$$I(x)=e^{\int \tan xdx}=e^{\ln (|\sec x|)}=\sec x$$

Multiply the linear equation by the integrating factor:

$$\frac{d}{dx}(v\cdot \sec x)=\sec x\cdot \sec x={\sec }^{2}x$$

Integrate both sides with respect to $x$:

$$v\sec x=\int {\sec }^{2}xdx+C$$

$$v\sec x=\tan x+C$$

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Step 3: Substitute Back and Finalize the Solution

Substitute $v=\frac{1}{y}$ back into the equation:

$$\frac{1}{y}\sec x=\tan x+C$$

Solve for $y$:

$$y=\frac{\sec x}{\tan x+C}$$

This can be further simplified using trigonometric identities ($\sec x=\frac{1}{\cos x}$, $\tan x=\frac{\sin x}{\cos x}$):

$$y=\frac{\frac{1}{\cos x}}{\frac{\sin x}{\cos x}+C\frac{\cos x}{\cos x}}=\frac{\frac{1}{\cos x}}{\frac{\sin x+C\cos x}{\cos x}}=\frac{1}{\sin x+C\cos x}$$

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Answer: The general solution to the differential equation is $y=\frac{1}{\sin x+C\cos x}$.