How do I patch together solutions for a discontinuous coefficient problem?

[cbarker1]

Discontinuous Coefficients Problem

Hello,

I need some help for part d.

The book's problem stated, "Occasions arise when the coefficient P(x) in a linear equation fails to be continuous because of the jump discontinuous. Fortunately, we may still obtain a 'reasonable' solution. For example consider the initial value problem

$\d{y}{x}+P(x)y=x$ , $y(0)=1$ where
$P(x)=\left\{\begin{array}{cc}1,&\mbox{ if }
0\le x\le2 \\3, & \mbox{ if } x>2\end{array}\right.$

a) Find the general solution for0\le x\le2$
b) Choose the constant in the solution of part a so that the initial condition is satisfied.
c.) Find the general solution $x\gt2$
d.) Now choose the constant in the general solution from part c so that the solution from part b and solution from part c agree $x=2$. By patching the two solutions together, we can obtain a continuous function that satisfies the differential equation except at x=2, where its derivative is undefined.

Work for part a:
$\d{y}{x}+P(x)y=x$ , $y(0)=1$ where
$P(x)=\left\{\begin{array}{cc}1,&\mbox{ if }
0\le x\le2 \\3, & \mbox{ if } x>2\end{array}\right.$

Since the interval is $0\le x \le 2$, I have to plugin in 1 for the P.

$\d{y}{x}+y=x$

Integrating Factor
$\mu(x)=e^{\int dx}=e^{x}$

$e^{x}\d{y}{x}+e^{x}y=xe^{x}$

$\left[e^{x}y\right]^{\prime}=xe^{x}$

$\int\left[e^{x}y\right]^{\prime}=\int xe^{x}dx$

Integrating by parts

$e^{x}y=e^{x}\left(x-1\right)+C$

$y=x-1+Ce^{-x}$

Work for part b.
$y(x)=x-1+Ce^{-x}$

$y(0)=1=-1+C$

$C=2$

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

Work for part C.

$\d{y}{x}+P(x)y=x$ , $y(0)=1$ where
$P(x)=\left\{\begin{array}{cc}1,&\mbox{ if }
0\le x\le2 \\3, & \mbox{ if } x>2\end{array}\right.$

Since the interval is $x \gt 2$, I have to plugin in 3 for the P.

$\d{y}{x}+3y=x$

Integrating Factor
$\mu(x)=e^{3\int dx}=e^{3x}$

$e^{3x}\d{y}{x}+3e^{3x}y=xe^{3x}$

$\left[e^{3x}y\right]^{\prime}=xe^{3x}$

$\int\left[e^{3x}y\right]^{\prime}=\int xe^{3x}dx$

Integrating by parts

$e^{3x}y=\frac{1}{9}e^{3x}\left(x-1\right)+C$

$y=\frac{1}{9}(3x-1)+Ce^{-3x}$

I am lost for part d. I do not know how to start with part d.

Thank you for your help.

Cbarker1

 

[cbarker1]

Since part (d says that x=2, plug in the value of x in for the answer from part b and part c and equality them too?