Question

Solve the Initial Value Problem:

\[y^\prime+2y=g(x),\qquad y(0)=0\]

\[g(x)=\left\{\begin{array}{ccc} 1, & \quad0\leq x\leq1 \\ 0, & \quad x>1 \end{array}\right.\]

Answer 1

\[\text{---------}\]\[\text{case 1 }\qquad 0\leq x\leq1 \]\[g(x)=1\]\[y^\prime+2y=1\]\[\frac{\text dy}{\text dx}=1-2y\]\[\int\frac{\text dy}{1-2y}=\int\text dx\]\[\frac{-1}{2} \int\frac{-2}{1-2y}=x+c\]\[\ln|1-2y|=-2x-2c\]\[1-2y=e^{-2c}e^{-2x}\]\[y=\frac{1-e^{-2c}e^{-2x}}2\]\[y(x)=Ae^{-2x}+\frac 12 \]\[y(0)=A+\frac 12=0\]\[A=-\frac 12\]\[y(x)=\frac 12 -\frac{e^{-2x}}2 \]

Answer 2

\[\text{---------}\]\[\text{case 2 } \quad x>1\]\[g(x)=0\]\[y^\prime+2y=0\]\[\frac{\text dy}{\text dx}=-2y\]\[\int\frac {\text dy}{y} =-2\int\text dx\]\[\ln |y|=-2x+c_1\]\[y(x)=e^{c_1}e^{-2x}=Ae^{-2x}\]\[y(0)=A=0\]\[A=0\]\[y(x)=0\]

Answer 3

how am i ment to write the final solution?
\[y(x)=\left\{\begin{array}{ccc} \frac 12 -\frac{e^{-2x}}2 , & \quad0\leq x\leq1 \\ 0, & \quad x>1 \end{array}\right.\]

Answer 4

there has to be a neater final solution

Answer 5

maybe like this?\[y(x)g(x)=\frac 12 -\frac{e^{-2x}}2\]

Answer 6

\[y=(\frac{1}{2}-\frac{e^{-3x}}{2})g(x)\]or u can say\[y=\frac{1}{2}-\frac{e^{-3x}}{2} , 0 \le x \le 1\]