Differential Equations: Speed-Solving Techniques (8)

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Tags Differential Equations

Question

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

Solution — Step by Step

This is a linear first-order ODE: dydx+P(x)y=Q(x)\dfrac{dy}{dx} + P(x)y = Q(x) with P=tan⁡xP = \tan x, Q=sin⁡xQ = \sin x.

IF=e∫P dx=e∫tan⁡x dx=eln⁡∣sec⁡x∣=sec⁡xIF = e^{\int P\,dx} = e^{\int \tan x\,dx} = e^{\ln|\sec x|} = \sec x

sec⁡xdydx+ysec⁡xtan⁡x=sec⁡xsin⁡x=tan⁡x\sec x \frac{dy}{dx} + y\sec x \tan x = \sec x \sin x = \tan x

The left side is ddx(ysec⁡x)\dfrac{d}{dx}(y\sec x). So:

ddx(ysec⁡x)=tan⁡x\frac{d}{dx}(y\sec x) = \tan x

Integrate:

ysec⁡x=∫tan⁡x dx=ln⁡∣sec⁡x∣+Cy \sec x = \int \tan x\,dx = \ln|\sec x| + C

y=cos⁡x⋅(ln⁡∣sec⁡x∣+C)y = \cos x \cdot (\ln|\sec x| + C)

Final answer: y=cos⁡x⋅ln⁡∣sec⁡x∣+Ccos⁡xy = \cos x \cdot \ln|\sec x| + C\cos x

Why This Works

The integrating factor turns a non-exact equation into an exact one. After multiplying, the LHS becomes ddx(y⋅IF)\dfrac{d}{dx}(y \cdot IF), which integrates trivially.

The IF for a linear ODE in standard form is always e∫P dxe^{\int P\,dx} — memorise this and you can solve any first-order linear equation in 90 seconds.

Alternative Method

For this specific form, you could also try variation of parameters: solve the homogeneous equation dy/dx+ytan⁡x=0dy/dx + y\tan x = 0 first (separable, gives y=Ccos⁡xy = C\cos x), then guess y=u(x)cos⁡xy = u(x)\cos x and solve for uu. More work, same answer.

Common Mistake

Students forget the absolute value in ln⁡∣sec⁡x∣\ln|\sec x|. The IF is eln⁡∣sec⁡x∣=∣sec⁡x∣e^{\ln|\sec x|} = |\sec x|, but for problems on intervals where sec⁡x>0\sec x > 0 (like −π/2<x<π/2-\pi/2 < x < \pi/2), we drop the absolute value.

Also: a common error is to write the IF as sec⁡xtan⁡x\sec x \tan x (multiplying instead of exponentiating). Always exponentiate ∫P dx\int P\,dx.

If the equation isn’t already in dy/dx+Py=Qdy/dx + Py = Q form, divide through to get there first. JEE often disguises linear ODEs by hiding the dy/dxdy/dx coefficient.

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