Answer to Question #216778 in Differential Equations for hwc

Question

The indicated function y₁(x)

 is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution y₂(x)

 of the homogeneous equation and a particular solution y_p(x)

 of the given nonhomogeneous equation.

y'' + y' = 1;    y₁ = 1

y₂(x)

= y_p(x)

= 

Answer 1

Solution

$y'' + y' = 1$

${d^2y\over dx^2}+{dy\over dx}=1$

First, reduce the differential equation to first order

Let $z={dy\over dx} \implies {dz\over dx}={d^2y\over dx^2}$

Substituting to differential equation, we have

${dz\over dx}+z=1$

${dz\over dx}=1-z$

Do reciprocal on both sides of the differential equation

${1\over 1-z}={dx\over dz}$

Multiply both sides of the differential equation by $dz$ to get

${dz\over 1-z}=dx$

Integrate both sides

$\int{dz\over 1-z}=\int dx$

$\implies -ln(1-z)=x+C$

$\implies ln(1-z)=-x+C$

Introduce exponential on both sides

$e^{ln(1-z)}=e^{-x+c}$

$\implies 1-z=e^{-x+C} \implies z=1-e^{-x+C}=1-Ce^{-x}$

But $z={dy\over dx}$

$\implies {dy\over dx}=1-Ce^{-x}$

Multiply both sides of the differential equation by $dx$

$\implies dy=(1-Ce^{-x})dx$

Integrate both sides

$\int dy=\int (1-Ce^{-x})dx$

$\implies y=x+C_1e^{-x}+C_2$

$\therefore y_2(x)=e^{-x}$ and $y_p(x)=x$