Question

The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section 4.2, y2 = y1(x) e−∫P(x) dx y 2 1 (x) dx (5) as instructed, to find a second solution y2(x). y'' + 2y' + y = 0; y1 = xe−x

Answer 1

First confirm that $y_1=xe^{-x}$ is a solution to the ODE,

$y''+2y'+y=0$

We have

${y_1}'=e^{-x}-xe^{-x}=(1-x)e^{-x}$

${y_1}''=-e^{-x}-(1-x)e^{-x}=(-2+x)e^{-x}$

Substituting into the ODE gives

$(-2+x)e^{-x}+2(1-x)e^{-x}+xe^{-x}=0$

Suppose $y_2(x)=v(x)y_1(x)$ is another solution to this ODE. Then

${y_2}'=v'y_1+v{y_1}'$

${y_2}''=v''y_1+2v'{y_1}'+v{y_1}''$

and substituting these into the ODE yields

$(v''y_1+2v'{y_1}'+v{y_1}'')+2(v'y_1+v{y_1}')+vy_1=0$

$xe^{-x}v''+2e^{-x}v'=0$

$xv''+2v'=0$

Let $w(x)=v'(x)$. Then the remaining ODE is linear in $w$:

$xw'+2w=0$

Multiply both sides by the integrating factor, $x$, and condense the left hand side as a derivative of a product:

$x^2w'+2xw=(x^2w)'=0$

Integrate both sides with respect to $x$ and solve for $w$:

$x^2w=C_1\implies w=C_1x^{-2}$

Back-substitute and integrate both sides with respect to $x$ to solve for $v$:

$v'=C_1x^{-2}\implies v=-C_1x^{-1}+C_2$

Back-substitute again to solve for $y_2$:

$\dfrac{y_2}{y_1}=C_2-\dfrac{C_1}x$

$\implies y_2=C_2xe^{-x}-C_1e^{-x}$

$y_1$ already captures the solution $xe^{-x}$, so the remaining one is

$\boxed{y_2=e^{-x}}$

Answer 2

A differential equation shows the relationship between functions and their derivatives.

The equation of $y_2(x)$ is: $y_2 = -e^{-x}$

The given parameters are:

$y_2 = y_1(x) \int\frac{ e^{(\int -P(x)\ dx)} }{ y_1^2(x) }dx$

$y" + 2y' + y = 0$

$y_1 = xe^{-x}$

The general equation is:

$y" + P(x) y' + Q(x)y = 0$

Compare the above equation to $y" + 2y' + y = 0$

$P(x) = 2$

Integrate:

$\int\limits^x_0 P(x') dx'= \int\limits^x_0 2 dx'$

$\int\limits^x_0 P(x') dx'= 2x|\limits^x_0$

$\int\limits^x_0 P(x') dx'= 2[x - 0]$

$\int\limits^x_0 P(x') dx'= 2[x ]$

$\int\limits^x_0 P(x') dx'= 2x$

We have:

$y_2 = y_1(x) \int\frac{ e^{(\int -P(x)\ dx)} }{ y_1^2(x) }dx$

The above equation becomes:

$y_2 = y_1(x) \int\frac{e^{(\int -2 dx)} }{ y_1^2(x) }dx$

Substitute $y_1 = xe^{-x}$

$y_2 = xe^{-x} \int\frac{ e^{(\int -2 dx)} }{ (xe^{-x})^2 }dx$

Integrate

$y_2 = xe^{-x} \int\frac{ e^{-2x}}{ (xe^{-x})^2 }dx$

Evaluate the exponents

$y_2 = xe^{-x} \int\frac{ e^{-2x}}{ x^2e^{-2x} }dx$

Cancel out common factors

$y_2 = xe^{-x} \int\frac{1}{ x^2 }dx$

Rewrite as:

$y_2 = xe^{-x} \int x^{-2}dx$

Integrate

$y_2 = xe^{-x} \times -\frac{1}{x} + c$

$y_2 = -e^{-x} + c$

Set c to 0.

$y_2 = -e^{-x} + 0$

$y_2 = -e^{-x}$

Hence, the equation of $y_2(x)$ is:

$y_2 = -e^{-x}$

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