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Given the first IVP $$y'+2y=\left\{\begin{matrix} 2 ,& 0 \le x <1\\ -2,& x \ge1 \end{matrix}\right.$$

With initial condition $y(0)=2$

1) Find the explicit solution on the interval $0 \le x <1$

2) Find the explicit solution on the interval $x \ge 1$

For 1) Since I.F =$e^{2x}$

hence $y.e^{2x}=\int ^1_0 2 e^{2x}dx+c\\ y=\frac{e^{2x}+c}{e^{2x}}\\ y=\frac{e^{2x}+1}{e^{2x}}$ for $y(0)=1$

how can we solve (2)

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    $\begingroup$ Why don't you use the Laplace transform. $\endgroup$ Commented Mar 13, 2018 at 1:43
  • $\begingroup$ You will have the same $c$ but with a different particular solution. $\endgroup$ Commented Mar 13, 2018 at 1:52

2 Answers 2

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You will have the same $c$ as it is about the homogeneous part of the solution, but with a different particular solution ($y_p$).

Here is a classic but general method of solving:

  • We first find the homogeneous solution of the ODE as below:

$$y'+2y=0$$

$\displaystyle \to \dfrac{y'}{y}=-2 \to \ln(y)=-2x+c \to y_h=ce^{-2x}$

  • For the particular solution we have:

$$y'+2y=2u(t)-4u(t-1)$$

Now by the method of undetermined coefficients, we conclude that:

$$y_p=\left\{\begin{matrix} 1 ,& 0 \le x <1\\ -1,& x \ge1 \end{matrix}\right.$$

  • The general solution of ODE is of the form: $y=y_p + y_h$, so:

$$y=\left\{\begin{matrix} ce^{-2x}+1 ,& 0 \le x <1\\ ce^{-2x}-1,& x \ge1 \end{matrix}\right.$$

For the first part, we know that $y(0)=2$, thus $c+1=2 \to c=1$, hence: $$y=\left\{\begin{matrix} e^{-2x}+1 ,& 0 \le x <1\\ e^{-2x}-1,& x \ge1 \end{matrix}\right.$$

Or simply:

$$\displaystyle y=e^{-2x}+u(t)-2u(t-1)$$

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$$y'+2y=\left\{\begin{matrix} 2 ,& 0 \le x <1\\ -2,& x \ge1 \end{matrix}\right.$$ 1) for $0 \le x <1$ $$y'+2y=2$$ $$(ye^{2x})'=2e^{2x}$$ $$y=2e^{-2x}\int e^{2x}dx=2e^{-2x}(\frac {e^{2x}}2+K_1)=K_1e^{-2x}+1$$ $$y(0)=2 \implies K_1+1=2 \implies K_1=1 \implies y=e^{-2x}+1$$ 2) for $ x \ge 1$ $$y'+2y=-2$$ $$y'=-2(y+1)$$ $$\int \frac {dy}{(y+1)}=-2x+K_2$$ $$\ln{(y+1)}=-2x+K_2$$ $$y+1=K_2e^{-2x}$$ $$y=K_2e^{-2x}-1$$

For continuity reason we must have at $x=1$ $$K_2e^{-2x}-1=e^{-2x}+1 \implies K_2=1+2e^2$$ $$y=(1+2e^2)e^{-2x}-1$$ Therefore $$\boxed{y(x)=\left\{ \begin{matrix} e^{-2x}+1 & 0 \le x < 1 \\ (1+2e^2)e^{-2x}-1 & x \ge 1 \end{matrix} \right.} $$

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