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Question: Solve the differential equation: \(\dfrac{dy}{dx}+2y\tan x=\sin x\)...

Solve the differential equation:
dydx+2ytan⁡x=sin⁡x\dfrac{dy}{dx}+2y\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+2ytan⁡x=sin⁡x\dfrac{dy}{dx}+2y\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)=2tan⁡xP\left( x \right)=2\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∫2tan⁡xdxIF={{e}^{\int{2\tan xdx}}}.
We know that ∫tan⁡xdx=ln⁡(sec⁡x)+C\int{\tan xdx}=\ln \left( \sec x \right)+C
Hence, we have IF=e2ln⁡(sec⁡x)=eln⁡(sec⁡2x)IF={{e}^{2\ln \left( \sec x \right)}}={{e}^{\ln \left( {{\sec }^{2}}x \right)}}
We know that eln⁡x=x{{e}^{\ln x}}=x
Hence, we have
IF=sec⁡2xIF={{\sec }^{2}}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⁡2x=∫sin⁡xsec⁡2xdxy{{\sec }^{2}}x=\int{\sin x{{\sec }^{2}}xdx}
Now, we have
sin⁡xsec⁡2x=sin⁡xsec⁡xsec⁡x=sin⁡xcos⁡xsec⁡x=sec⁡xtan⁡x\sin x{{\sec }^{2}}x=\sin x\sec x\sec x=\dfrac{\sin x}{\cos x}\sec x=\sec x\tan x
Hence, we have
ysec⁡2x=∫sec⁡xtan⁡xdxy{{\sec }^{2}}x=\int{\sec x\tan xdx}
We know that
∫sec⁡xtan⁡xdx=sec⁡x\int{\sec x\tan xdx}=\sec x
Hence, we have
ysec⁡2x=sec⁡x+Cy{{\sec }^{2}}x=\sec x+C
Dividing both sides by sec⁡2x{{\sec }^{2}}x, we get
y=cos⁡x+Ccos⁡2xy=\cos x+C{{\cos }^{2}}x, which is the required solution of the given differential equation.

Note: Verification.
We have
y=cos⁡x+Ccos⁡2xy=\cos x+C{{\cos }^{2}}x
Differentiating both sides, we get
dydx=−sin⁡x−2Csin⁡xcos⁡x\dfrac{dy}{dx}=-\sin x-2C\sin x\cos x
Adding 2ytan⁡x2y\tan x on both sides, we get
dydx+2ytan⁡x=−sin⁡x−2Csin⁡xcos⁡x+2tan⁡x(cos⁡x+Ccos⁡2x)\dfrac{dy}{dx}+2y\tan x=-\sin x-2C\sin x\cos x+2\tan x\left( \cos x+C{{\cos }^{2}}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+2ytan⁡x=−sin⁡x−2Csin⁡xcos⁡x+2sin⁡x+2Csin⁡xcos⁡x\dfrac{dy}{dx}+2y\tan x=-\sin x-2C\sin x\cos x+2\sin x+2C\sin x\cos x
Simplifying, we get
dydx+2ytan⁡x=sin⁡x\dfrac{dy}{dx}+2y\tan x=\sin x, which is the given differential equation.
Hence our answer is verified to be correct.