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Solve the differential equation: ( D 2 + 4 ) y = tan ( 2 x )

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Concepts

Linear differential equations with constant coefficients, Complementary function, Particular integral, Operator method, Integration by parts

Explanation

We are given the differential equation: (D2+4)y=tan(2x) where D=dxd​.

We solve such equations by finding:

  1. The complementary function (CF) for the homogeneous part (D2+4)y=0.
  2. The particular integral (PI) for the non-homogeneous part tan(2x).

Step-By-Step Solution

Step 1

Find the Complementary Function (CF): Solve (D2+4)y=0

The auxiliary equation is: m2+4=0 m2=−4 m=±2i

So, the CF is: yCF​=C1​cos(2x)+C2​sin(2x)

Step 2

Find the Particular Integral (PI): The PI is given by: yPI​=D2+41​tan(2x)

This is not a standard form, so we use the operator method:

Let f(x)=tan(2x). We need to find yPI​ such that: (D2+4)yPI​=tan(2x)

Let us use the method of variation of parameters.

The complementary solution is based on y1​=cos(2x) and y2​=sin(2x).

The particular solution is: yPI​=−y1​∫Wy2​⋅f(x)​dx+y2​∫Wy1​⋅f(x)​dx where W is the Wronskian of y1​ and y2​.

Calculate the Wronskian: W=∣∣​cos(2x)−2sin(2x)​sin(2x)2cos(2x)​∣∣​=2cos2(2x)+2sin2(2x)=2

So, yPI​=−cos(2x)∫2sin(2x)tan(2x)​dx+sin(2x)∫2cos(2x)tan(2x)​dx

Recall tan(2x)=cos(2x)sin(2x)​

So,

  • sin(2x)tan(2x)=sin(2x)⋅cos(2x)sin(2x)​=cos(2x)sin2(2x)​
  • cos(2x)tan(2x)=cos(2x)⋅cos(2x)sin(2x)​=sin(2x)

So, yPI​=−cos(2x)∫2cos(2x)sin2(2x)​dx+sin(2x)∫2sin(2x)​dx

Let us compute each integral separately.

Integral 1:

∫2cos(2x)sin2(2x)​dx

Recall sin2(2x)=1−cos2(2x), but let’s use sin2(2x)=21−cos(4x)​.

So, ∫2cos(2x)sin2(2x)​dx=∫4cos(2x)1−cos(4x)​dx Split the integral: =41​∫cos(2x)1​dx−41​∫cos(2x)cos(4x)​dx

Now,

  • ∫cos(2x)1​dx=∫sec(2x)dx=21​ln∣sec(2x)+tan(2x)∣
  • cos(2x)cos(4x)​=2cos2(2x)−1 (using cos(4x)=2cos2(2x)−1)

So, ∫cos(2x)cos(4x)​dx=∫[2cos(2x)−sec(2x)]dx=2∫cos(2x)dx−∫sec(2x)dx

  • ∫cos(2x)dx=21​sin(2x)
  • ∫sec(2x)dx=21​ln∣sec(2x)+tan(2x)∣

So, ∫cos(2x)cos(4x)​dx=2⋅21​sin(2x)−21​ln∣sec(2x)+tan(2x)∣=sin(2x)−21​ln∣sec(2x)+tan(2x)∣

Now, plug back into the split integral: 41​∫cos(2x)1​dx−41​∫cos(2x)cos(4x)​dx=41​⋅21​ln∣sec(2x)+tan(2x)∣−41​[sin(2x)−21​ln∣sec(2x)+tan(2x)∣] =81​ln∣sec(2x)+tan(2x)∣−41​sin(2x)+81​ln∣sec(2x)+tan(2x)∣ =41​ln∣sec(2x)+tan(2x)∣−41​sin(2x)

Integral 2:

∫2sin(2x)​dx=21​∫sin(2x)dx=−41​cos(2x)

Step 3

Substitute back into yPI​:

Recall: yPI​=−cos(2x)[41​ln∣sec(2x)+tan(2x)∣−41​sin(2x)]+sin(2x)[−41​cos(2x)]

Expand: yPI​=−41​cos(2x)ln∣sec(2x)+tan(2x)∣+41​cos(2x)sin(2x)−41​sin(2x)cos(2x)

Notice that cos(2x)sin(2x)−sin(2x)cos(2x)=0.

So, yPI​=−41​cos(2x)ln∣sec(2x)+tan(2x)∣

Step 4

Write the General Solution:

Combine CF and PI: y(x)=C1​cos(2x)+C2​sin(2x)−41​cos(2x)ln∣sec(2x)+tan(2x)∣

Final Answer

y(x)=C1​cos(2x)+C2​sin(2x)−41​cos(2x)ln∣sec(2x)+tan(2x)∣ where C1​ and C2​ are arbitrary constants.