What is the general solution of the differential equation? # cosy(ln(secx+tanx))dx=cosx(ln(secy+tany))dy #
# sec^2y = sec^2x + C #
We have:
This ODE can be rearranged from differential form to standard form as follows:
Now that this ODE can be separated, we can "separate the variables" to obtain:
Think about the RHS integral.
We are able to substitute:
This is what we obtain when we replace it in the integral:
We can now integrate both sides of [B] using this result to obtain:
That is [A]'s General Solution.
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To find the general solution of the given differential equation, ( \cos y \ln(\sec x + \tan x) , dx = \cos x \ln(\sec y + \tan y) , dy ), you can separate the variables and integrate both sides. After integration, you'll obtain the solution in implicit form. However, due to the complexity of the equation, obtaining an explicit solution may not be feasible.
The steps to solve the differential equation are as follows:
- Separate the variables.
- Integrate both sides.
- Solve for any constants of integration.
- Simplify the equation as much as possible.
Following these steps will lead to the general solution of the differential equation.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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