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Question: Solve the differential equation \(\dfrac{dy}{dx}+y\tan x=\sin x\)...

Solve the differential equation
dydx+ytan⁡x=sin⁡x\dfrac{dy}{dx}+y\tan x=\sin x

Explanation

Solution

Hint: Observe that the given differential equation is a linear differential equation, i.e. a differential equation of the form dydx+P(x)y=Q(x)\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right). The solution of a linear differential equation is given by
yI.F=∫Q(x)I.FdxyI.F=\int{Q\left( x \right)I.Fdx}, where IF=e∫P(x)dxIF={{e}^{\int{P\left( x \right)dx}}}. Find the integrating factor of the given differential equation and hence find the solution of the given equation. Verify your answer.

Complete step-by-step answer:
We have dydx+ytan⁡x=sin⁡x\dfrac{dy}{dx}+y\tan x=\sin x, which is of the form of dydx+P(x)y=Q(x)\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right).
Here P(x)=tan⁡xP\left( x \right)=\tan x and Q(x)=sin⁡xQ\left( x \right)=\sin x.
Now, we know that the integrating factor IF of a linear differential equation is given by
IF=e∫P(x)dxIF={{e}^{\int{P\left( x \right)dx}}}
Hence, the integrating factor of the given differential equation is given by
IF=e∫tan⁡xdxIF={{e}^{\int{\tan xdx}}}.
We know that ∫tan⁡xdx=ln⁡(sec⁡x)+C\int{\tan xdx}=\ln \left( \sec x \right)+C
Hence, we have IF=eln⁡(sec⁡x)IF={{e}^{\ln \left( \sec x \right)}}
We know that eln⁡x=x{{e}^{\ln x}}=x
Hence, we have
IF=sec⁡xIF=\sec x
Now, we know that the solution of a linear differential equation is given by
yIF=∫Q(x)IFdxyIF=\int{Q\left( x \right)IFdx}
Hence, the solution of the given equation is
ysec⁡x=∫sin⁡xsec⁡xdxy\sec x=\int{\sin x\sec xdx}
Now, we have
sin⁡xsec⁡x=sin⁡xcos⁡x=tan⁡x\sin x\sec x=\dfrac{\sin x}{\cos x}=\tan x
Hence, we have
ysec⁡x=∫tan⁡xdxy\sec x=\int{\tan xdx}
We know that
∫tan⁡xdx=ln⁡(sec⁡x)\int{\tan xdx}=\ln \left( \sec x \right)
Hence, we have
ysec⁡x=ln⁡(sec⁡x)+Cy\sec x=\ln \left( \sec x \right)+C
Dividing both sides by sec⁡x\sec x, we get
y=cos⁡xln⁡(sec⁡x)+Ccos⁡xy=\cos x\ln \left( \sec x \right)+C\cos x, which is the required solution of the given differential equation.

Note: Verification.
We have
y=cos⁡xln⁡(sec⁡x)+Ccos⁡xy=\cos x\ln \left( \sec x \right)+C\cos x
Differentiating both sides, we get
dydx=−sin⁡xln⁡(sec⁡x)+cos⁡xtan⁡x−Csin⁡x\dfrac{dy}{dx}=-\sin x\ln \left( \sec x \right)+\cos x\tan x-C\sin x
Adding ytan⁡xy\tan x on both sides, we get
dydx+ytan⁡x=dydx=−sin⁡xln⁡(sec⁡x)+cos⁡xtan⁡x−Csin⁡x+tan⁡x(cos⁡xln⁡(sec⁡x)+Ccos⁡x)\dfrac{dy}{dx}+y\tan x=\dfrac{dy}{dx}=-\sin x\ln \left( \sec x \right)+\cos x\tan x-C\sin x+\tan x\left( \cos x\ln \left( \sec x \right)+C\cos x \right)
We have tan⁡xcos⁡x=sin⁡xcos⁡xcos⁡x=sin⁡x\tan x\cos x=\dfrac{\sin x}{\cos x}\cos x=\sin x
Hence, we have
dydx+ytan⁡x=−sin⁡xln⁡(sec⁡x)+sin⁡x−Csin⁡x+sin⁡xln⁡(sec⁡x)+Csin⁡x\dfrac{dy}{dx}+y\tan x=-\sin x\ln \left( \sec x \right)+\sin x-C\sin x+\sin x\ln \left( \sec x \right)+C\sin x
Simplifying, we get
dydx+ytan⁡x=sin⁡x\dfrac{dy}{dx}+y\tan x=\sin x, which is the given differential equation.
Hence our answer is verified to be correct.