Question

solve the differential equation cosxdy/dx+4ysinx=4√y secx​

Answer 1

Step-by-step explanation:

We have,

$\cos(x) \frac{dy}{dx} + 4y \sin(x) = 4 \sqrt{y} \sec(x)$

$\implies \frac{dy}{dx} + 4y \tan(x) = 4 \sqrt{y} \sec^{2} (x)$

$\implies \frac{1}{ \sqrt{y} } \frac{dy}{dx} + 4 \sqrt{y} \tan(x) = 4 \sec^{2} (x)$

$Let \: \: 2 \sqrt{y} = v \\ \implies \frac{1}{ \sqrt{y} } \frac{dy}{dx} = \frac{dv}{dx}$

Now,

$\implies \frac{dv}{dx} +2v \tan(x) = 4 \sec^{2} (x)$

$I.F = {e}^{ \int2 \tan(x) dx} = {e}^{2 \ln( \sec(x) ) } = {e}^{ \ ln( \sec^{2}(x) ) } = \sec^{2}(x)$

Required solution

$v .\sec^{2}(x) = \int4\sec^{2}(x) \sec^{2}(x)dx$

$\implies \: v .\sec^{2}(x) = 4 \int(1 + \tan^{2}(x)) \sec^{2}(x)dx$

On RHS, we put $\tan(x) = t\:\: \implies \sec^{2}dx=dt$

so,

$\implies \: v .\sec^{2}(x) = 4 \int(1 + t^{2}) dt$

$\implies \: v .\sec^{2}(x) = 4t + 4\frac{ t^{3}}{3} + C$

$\implies \: v .\sec^{2}(x) = 4t + \frac{4 }{3}t^{3}+ C$

$\implies \: v .\sec^{2}(x) = 4 \tan(x) + \frac{4 }{3} \tan^{3}(x)+ C$

$\implies \: 2 \sqrt{y} \sec^{2}(x) = 4 \tan(x) + \frac{4 }{3} \tan^{3}(x)+ C$

$\implies \: \sqrt{y} \sec^{2}(x) = 2\tan(x) + \frac{2 }{3} \tan^{3}(x)+ \frac{ C }{2}$

$\implies \: \sqrt{y} \sec^{2}(x) = 2\tan(x) + \frac{2 }{3} \tan^{3}(x)+ K$

Answer 2

Answer: The solution of the given differential equation cos x dy/dx + 4y sin x = 4y sec x be,

1/4 log y = tan x - log sec x + C.

Given the differential equation

cos x dy/dx + 4y sin x = 4y sec x

On simplifying the differential equation

cos x dy/dx = 4y sec x - 4y sin x

cos x dy/dx = 4y (sec x - sin x)

On cross-multiplying

1/4y dy = (sec x - sin x)/cos x dx

Now, on Integrating both the sides

∫1/4y dy = ∫(sec² x - tan x) dx

1/4∫1/y dy = ∫sec² x - ∫tan x dx

1/4 log y = tan x - log sec x + C

The solution of the given differential equation cos x dy/dx + 4y sin x = 4y sec x be,

1/4 log y = tan x - log sec x + C.

To conclude in one sentence, the solution of the given differential equation cos x dy/dx + 4y sin x = 4y sec x be,

1/4 log y = tan x - log sec x + C.

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