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Math  /  Calculus

QuestionH.w Solve: y′′+2y′+y=4e−xln⁡x\quad y^{\prime \prime}+2 y^{\prime}+y=4 e^{-x} \ln x

Studdy Solution

STEP 1

What is this asking? We're trying to find a function yy whose second derivative plus twice its first derivative plus itself equals 4e−xln⁡x4e^{-x}\ln x. Watch out! Don't forget to consider both the complementary solution and the particular solution!

STEP 2

1. Find the complementary solution
2. Find the particular solution
3. Combine solutions

STEP 3

Alright, let's **kick things off** with the **homogeneous equation**: y′′+2y′+y=0y^{\prime \prime} + 2y^{\prime} + y = 0.
We're looking for solutions of the form y=erxy = e^{rx}.

STEP 4

Substituting this into our equation gives us r2erx+2rerx+erx=0r^2 e^{rx} + 2re^{rx} + e^{rx} = 0.
Since erxe^{rx} is never zero, we can divide both sides by it, which leaves us with the **characteristic equation**: r2+2r+1=0r^2 + 2r + 1 = 0.

STEP 5

This is a **perfect square trinomial**, which factors to (r+1)2=0(r+1)^2 = 0.
This gives us a **repeated root** r=−1r = -1.

STEP 6

With a repeated root, our **complementary solution** is yc=c1e−x+c2xe−xy_c = c_1e^{-x} + c_2xe^{-x}.
Woohoo! One part down!

STEP 7

Now for the **particular solution**.
Since our right-hand side is 4e−xln⁡x4e^{-x}\ln x, we'll use **variation of parameters**.
Let yp=u1(x)e−x+u2(x)xe−xy_p = u_1(x)e^{-x} + u_2(x)xe^{-x}, where y1=e−xy_1 = e^{-x} and y2=xe−xy_2 = xe^{-x} are the solutions we found earlier.

STEP 8

The **Wronskian** of y1y_1 and y2y_2 is given by W(y1,y2)=∣e−xxe−x−e−xe−x−xe−x∣=e−2x. W(y_1, y_2) = \begin{vmatrix} e^{-x} & xe^{-x} \\ -e^{-x} & e^{-x} - xe^{-x} \end{vmatrix} = e^{-2x}.

STEP 9

Now we can find u1u_1 and u2u_2: u1=−∫y2⋅4e−xln⁡xW(y1,y2)dx=−∫xe−x⋅4e−xln⁡xe−2xdx=−4∫xln⁡x dx u_1 = -\int \frac{y_2 \cdot 4e^{-x}\ln x}{W(y_1, y_2)} dx = -\int \frac{xe^{-x} \cdot 4e^{-x}\ln x}{e^{-2x}} dx = -4\int x\ln x \, dx Using integration by parts, we get u1=−2x2ln⁡x+x2u_1 = -2x^2\ln x + x^2.
u2=∫y1⋅4e−xln⁡xW(y1,y2)dx=∫e−x⋅4e−xln⁡xe−2xdx=4∫ln⁡x dx u_2 = \int \frac{y_1 \cdot 4e^{-x}\ln x}{W(y_1, y_2)} dx = \int \frac{e^{-x} \cdot 4e^{-x}\ln x}{e^{-2x}} dx = 4\int \ln x \, dx Again, using integration by parts, we find u2=4xln⁡x−4xu_2 = 4x\ln x - 4x.

STEP 10

So, our **particular solution** is yp=(−2x2ln⁡x+x2)e−x+(4xln⁡x−4x)xe−x=2x2e−xln⁡x−3x2e−xy_p = (-2x^2\ln x + x^2)e^{-x} + (4x\ln x - 4x)xe^{-x} = 2x^2e^{-x}\ln x - 3x^2e^{-x}.

STEP 11

Finally, we **combine** the complementary and particular solutions to get the **general solution**: y=yc+yp=c1e−x+c2xe−x+2x2e−xln⁡x−3x2e−x. y = y_c + y_p = c_1e^{-x} + c_2xe^{-x} + 2x^2e^{-x}\ln x - 3x^2e^{-x}.

STEP 12

The general solution to the differential equation y′′+2y′+y=4e−xln⁡xy^{\prime \prime}+2 y^{\prime}+y=4 e^{-x} \ln x is y=c1e−x+c2xe−x+2x2e−xln⁡x−3x2e−xy = c_1e^{-x} + c_2xe^{-x} + 2x^2e^{-x}\ln x - 3x^2e^{-x}.

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