Verify that the given function is a solution of the differential equation. y sec x = tan x + c;
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To verify that the given function is a solution of the differential equation. y sec x = tan x + c;
![\frac{dy}{dx}+y\tan x = \sec x\\ \frac{dy}{dx}+y\tan x = \sec x\\](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%2By%5Ctan+x+%3D+%5Csec+x%5C%5C)
calculate dy/dx from given equation,and convert
![y \: sec \: x = tan \: x + c \\ \\ y \: sec \: x \: tan \: x + sec \: x \frac{dy}{dx} = {sec}^{2} \: x \\ \\ y \: sec \: x = tan \: x + c \\ \\ y \: sec \: x \: tan \: x + sec \: x \frac{dy}{dx} = {sec}^{2} \: x \\ \\](https://tex.z-dn.net/?f=y+%5C%3A+sec+%5C%3A+x+%3D+tan+%5C%3A+x+%2B+c+%5C%5C+%5C%5C+y+%5C%3A+sec+%5C%3A+x+%5C%3A+tan+%5C%3A+x+%2B+sec+%5C%3A+x+%5Cfrac%7Bdy%7D%7Bdx%7D+%3D+%7Bsec%7D%5E%7B2%7D+%5C%3A+x+%5C%5C+%5C%5C+)
taking sec x common
![sec \: x \: (y \: tan \: x + \frac{dy}{dx}) = {sec}^{2} \: x \\ \\ (y \: tan \: x + \frac{dy}{dx}) = sec \: x \\ \\ sec \: x \: (y \: tan \: x + \frac{dy}{dx}) = {sec}^{2} \: x \\ \\ (y \: tan \: x + \frac{dy}{dx}) = sec \: x \\ \\](https://tex.z-dn.net/?f=sec+%5C%3A+x+%5C%3A+%28y+%5C%3A+tan+%5C%3A+x+%2B+%5Cfrac%7Bdy%7D%7Bdx%7D%29+%3D+%7Bsec%7D%5E%7B2%7D+%5C%3A+x+%5C%5C+%5C%5C+%28y+%5C%3A+tan+%5C%3A+x+%2B+%5Cfrac%7Bdy%7D%7Bdx%7D%29+%3D+sec+%5C%3A+x+%5C%5C+%5C%5C+)
is the desired equation.
calculate dy/dx from given equation,and convert
taking sec x common
is the desired equation.
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