丁香园AV

Find the general solution of the following homogeneous differential equations.

1. \(\ds y'+5y=0\)

   Answer
   \(\ds y=Ae^{-5t}\)
   Solution
   This is a first order homogeneous linear differential equation. Therefore, we separate variables:

   \begin{equation*} \begin{split} \diff{y}{t} \amp= -5y \\ \int \frac{1}{y}\,dy \amp= -5 \int \,dt \\ \ln |y| \amp= -5t + C \\ y(t) \amp= e^{-5t+C} = Ae^{-5t},\end{split} \end{equation*}

   for any constant \(A\text{.}\)

2. \(\ds y'-2y=0\)

   Answer
   \(\ds y=Ae^{2t}\)
   Solution
   This is a first order homogeneous linear differential equation. Therefore, we separate variables:

   \begin{equation*} \begin{split} \diff{y}{t} \amp= 2y \\ \int \frac{1}{y}\,dy \amp= 2 \int \,dt \\ \ln |y| \amp= 2t + C \\ y(t) \amp= e^{2t+C} = Ae^{2t},\end{split} \end{equation*}

   for any constant \(A\text{.}\)

3. \(\ds y'+{y\over 1+t^2}=0\)

   Answer
   \(\ds y=Ae^{-\arctan t}\)
   Solution
   This is a first order homogeneous linear differential equation. Therefore, we separate variables:

   \begin{equation*} \begin{split} \diff{y}{t} \amp= -\frac{y}{1+t^2} \\ \int \frac{1}{y}\,dy \amp= - \int \frac{1}{1+t^2} \,dt \\ \ln |y|\amp= -\tan^{-1}(t) + C \\ y(t) \amp= e^{-\tan^{-1}(t)+C} = Ae^{-\tan^{-1}(t)},\end{split} \end{equation*}

   for any constant \(A\text{.}\)

4. \(\ds y'+t^2y=0\)

   Answer
   \(\ds y=Ae^{-t^3/3}\)
   Solution
   This is a first order homogeneous linear differential equation. Therefore, we separate variables:

   \begin{equation*} \begin{split} \diff{y}{t} \amp= -t^2 y \\ \int \frac{1}{y}\,dy \amp= - \int t^2 \,dt \\ \ln |y| \amp= -\frac{t^3}{3} + C \\ y(t) \amp= e^{-\frac{t^3}{3}+C} = Ae^{-\frac{t^3}{3}},\end{split}s \end{equation*}

   for any constant \(A\text{.}\)

Solve the following initial value problems concerning homogeneous DEs.

1. \(\ds y' + y=0\text{,}\) \(y(0)=4\)

   Answer
   \(\ds y=4e^{-t}\)
   Solution
   We first find the general solution by separating variables. Notice that \(y=0\) is a constant solution, but is not a solution to the IVP.

   \begin{equation*} \begin{split} \diff{y}{t} \amp= -y \\ \int \frac{1}{y} \,dy \amp= -\int \,dt \\ \ln |y| \amp= -t + C\\ y(t) \amp= A e^{-t}. \end{split} \end{equation*}

   We now apply the initial condition:

   \begin{equation*} y(0) = 4 \implies A = 4. \end{equation*}

   Hence, the solution to the IVP is

   \begin{equation*} y(t) = 4e^{-t}. \end{equation*}

2. \(\ds y' -3y=0\text{,}\) \(y(1)=-2\)

   Answer
   \(\ds y=-2e^{3t-3}\)
   Solution
   We first find the general solution by separating variables. Notice that \(y=0\) is a constant solution, but is not a solution to the IVP.

   \begin{equation*} \begin{split} \diff{y}{t} \amp= 3y \\ \int \frac{1}{y} \,dy \amp= \int 3\,dt \\ \ln |y| \amp= 3t + C\\ y(t) \amp= A e^{3t}. \end{split} \end{equation*}

   We now apply the initial condition:

   \begin{equation*} y(1) = -2 \implies A = -2e^{-3}. \end{equation*}

   Hence, the solution to the IVP is

   \begin{equation*} y(t) = -2e^{3(t-1)}. \end{equation*}

3. \(\ds y' + y\sin t = 0\text{,}\) \(y(\pi)=1\)

   Answer
   \(\ds y=e^{1+\cos t}\)
   Solution
   We first find the general solution by separating variables:

   \begin{equation*} \begin{split} \diff{y}{t} \amp= -y\sin t \\ \int \frac{1}{y} \,dy \amp= -\int \sin t \,dt \\ \ln |y| \amp= \cos t + C\\ y(t) \amp= A e^{\cos t}. \end{split} \end{equation*}

   We now apply the initial condition:

   \begin{equation*} y(\pi) = 1 \implies A = \frac{1}{e^{-1}} = e. \end{equation*}

   Hence, the solution to the IVP is

   \begin{equation*} y(t) = e^{\cos(t) +1}. \end{equation*}

4. \(\ds y' +ye^t=0\text{,}\) \(y(0)=e\)

   Answer
   \(\ds y=e^2e^{-e^t}\)
   Solution
   We first find the general solution by separating variables:

   \begin{equation*} \begin{split} \diff{y}{t} \amp= -ye^t \\ \int \frac{1}{y} \,dy \amp= -\int e^t \,dt \\ \ln |y| \amp= -e^t + C\\ y(t) \amp= A e^{-e^t}. \end{split} \end{equation*}

   We now apply the initial condition:

   \begin{equation*} y(0) = e \implies A = \frac{e}{e^{-1}} = e^2. \end{equation*}

   Hence, the solution to the IVP is

   \begin{equation*} y(t) = e^{2-e^t}. \end{equation*}

5. \(\ds y' +y\sqrt{1+t^4}=0\text{,}\) \(y(0)=0\)

   Answer
   \(\ds y=0\)
   Solution
   We notice that \(y=0\) is a constant solution to the IVP.

6. \(\ds y' + y\cos(e^t)=0\text{,}\) \(y(0)=0\)

   Answer
   \(\ds y=0\)
   Solution
   We notice that \(y=0\) is a constant solution to the IVP.

7. \(\ds ty' - 2y = 0\text{,}\) \(y(1)=4\)

   Answer
   \(\ds y=4t^2\)
   Solution
   We first find the general solution by separating variables.

   \begin{equation*} \begin{split} t\diff{y}{t} \amp= 2y \\ \int \frac{1}{y} \,dy \amp= 2\int \frac{1}{t}\,dt \\ \ln |y| \amp= 2\ln|t| + C\\ y(t) \amp= e^{2\ln |t| + C}\\ \amp= Ae^{2\ln |t|}\\ \amp= A e^{\ln t^2}\\ \amp= At^2 \end{split} \end{equation*}

   We now apply the initial condition:

   \begin{equation*} y(1) = 4 \implies A = 4. \end{equation*}

   Hence, the solution to the IVP is

   \begin{equation*} y(t) = 4t^2. \end{equation*}

8. \(\ds t^2y' + y = 0\text{,}\) \(y(1)=-2\text{,}\) \(t>0\)

   Answer
   \(\ds y=-2e^{(1/t)-1}\)
   Solution
   We first find the general solution by separating variables.

   \begin{equation*} \begin{split} t^2\diff{y}{t} \amp= -y \\ \int \frac{1}{y} \,dy \amp= -\int \frac{1}{t^2}\,dt \\ \ln |y| \amp= \frac{1}{t} + C\\ y(t) \amp= e^{1/t + C}\\ \amp= Ae^{1/t} \end{split} \end{equation*}

   We now apply the initial condition:

   \begin{equation*} y(1) = -2 \implies A = -2e^{-1}. \end{equation*}

   Hence, the solution to the IVP is

   \begin{equation*} y(t) = -2e^{1/t-1}. \end{equation*}

9. \(\ds t^3y' = 2y\text{,}\) \(y(1)=1\text{,}\) \(t>0\)

   Answer
   \(\ds y=e^{1-t^{-2}}\)
   Solution
   We first find the general solution by separating variables.

   \begin{equation*} \begin{split} t^3\diff{y}{t} \amp=2y \\ \int \frac{1}{y} \,dy \amp= 2\int \frac{1}{t^3}\,dt \\ \ln |y| \amp= -2\frac{1}{2t^2} + C\\ y(t) \amp= e^{-1/(t^2) + C}\\ \amp= Ae^{-1/t^2} \end{split} \end{equation*}

   We now apply the initial condition:

   \begin{equation*} y(1) = 1 \implies A = e^{1}. \end{equation*}

   Hence, the solution to the IVP is

   \begin{equation*} y(t) = e^{1-1/t^2}. \end{equation*}

10. \(\ds t^3y' = 2y\text{,}\) \(y(1)=0\text{,}\) \(t>0\)

    Answer
    \(\ds y=0\)
    Solution
    We first find the general solution by separating variables.

    \begin{equation*} \begin{split} t^3\diff{y}{t} \amp =2y \\ \int \frac{1}{y} \,dy \amp = 2\int \frac{1}{t^3}\,dt \\ \ln |y| \amp = -2\frac{1}{2t^2} + C\\ y(t) \amp = e^{-1/(t^2) + C}\\ \amp= Ae^{-1/t^2} \end{split} \end{equation*}

    We now apply the initial condition:

    \begin{equation*} y(1) = 0 \implies A = 0. \end{equation*}

    Hence, the solution to the IVP is

    \begin{equation*} y(t) = 0, \end{equation*}

    but in order to separate variables we had to assume that \(y\neq 0\text{.}\) However, by inspection, we notice that \(y(t) = 0 \) is indeed a solution to this IVP.

Find the general solution of the following non-homogeneous differential equations.

1. \(\ds y' +4y=8\)

   Answer
   \(\ds y=Ae^{-4t}+2\)
   Solution
   We first solve the corresponding homogeneous equation using separation of variables:

   \begin{equation*} \begin{split} \diff{y}{t} \amp= -4y \\ \int \frac{1}{y}\,dy \amp= -4\int \,dt \\ \ln|y| \amp= -4t+ C\\ y(t) \amp= Ae^{-4t}.\end{split} \end{equation*}

   We now notice that \(y=2\) is a particular solution to the DE:

   \begin{equation*} \diff{}{t} (2) + 4(2) = 8. \end{equation*}

   Hence, the general solution is

   \begin{equation*} y(t) = Ae^{-4t} + 2. \end{equation*}

2. \(\ds y'-2y=6\)

   Answer
   \(\ds y=Ae^{2t}-3\)
   Solution
   We first solve the corresponding homogeneous equation using separation of variables:

   \begin{equation*} \begin{split} \diff{y}{t} \amp= 2y \\ \int \frac{1}{y}\,dy \amp= 2\int \,dt \\ \ln|y| \amp= 2t+ C\\ y(t) \amp= Ae^{2t}.\end{split} \end{equation*}

   We now notice that \(y=-3\) is a particular solution to the DE:

   \begin{equation*} \diff{}{t} (-3) -2(-3) = 6. \end{equation*}

   Hence, the general solution is

   \begin{equation*} y(t) = Ae^{2t} -3.\ \end{equation*}

3. \(\ds y' +ty=5t\)

   Answer
   \(\ds y=Ae^{-(1/2)t^2}+5\)
   Solution
   We first solve the corresponding homogeneous equation using separation of variables:

   \begin{equation*} \begin{split} \diff{y}{t} \amp= -ty \\ \int \frac{1}{y}\,dy \amp= -\int t \,dt \\ \ln|y| \amp= -\frac{t^2}{2} + C\\ y(t) \amp= Ae^{-t^2/2}.\end{split} \end{equation*}

   We now notice that \(y=5\) is a particular solution to the DE:

   \begin{equation*} \diff{}{t} (5) + t(5) = 5t. \end{equation*}

   Hence, the general solution is

   \begin{equation*} y(t) = Ae^{-t^2/2} + 5. \end{equation*}

4. \(\ds y'+e^ty=-2e^t\)

   Answer
   \(\ds y=Ae^{-e^t}-2\)
   Solution
   We first solve the corresponding homogeneous equation using separation of variables:

   \begin{equation*} \begin{split} \diff{y}{t} \amp=-e^t y \\ \int \frac{1}{y}\,dy \amp= -\int e^t \,dt \\ \ln|y| \amp= -e^t + C\\ y(t) \amp= Ae^{-e^t}.\end{split} \end{equation*}

   We now notice that \(y=-2\) is a particular solution to the DE:

   \begin{equation*} \diff{}{t} (-2) + e^t (-2) = -2e^t. \end{equation*}

   Hence, the general solution is

   \begin{equation*} y(t) = Ae^{-e^t} -2 . \end{equation*}

5. \(\ds y'-y=t^2\)

   Answer
   \(\ds y=Ae^{t}-t^2-2t-2\)
   Solution
   We first solve the corresponding homogeneous equation using separation of variables:

   \begin{equation*} \begin{split} \diff{y}{t} \amp=y \\ \int \frac{1}{y}\,dy \amp= \int \,dt \\ \ln|y| \amp= t + C\\ y(t) \amp= Ae^{t}.\end{split} \end{equation*}

   It is difficult to guess a particular solution to this DE (notice that there are no constant solutions which satisfy the DE). Therefore, we use the method of variation of parameters: since the right-hand side is a quadratic polynomial, we guess that a particular solution is of the form \(y = at^2 +bt + c\text{.}\) Now apply the DE:

   \begin{equation*} y' - y = 2at + b - (at^2+bt+c) = -at^2 +(2a-b)t + b-c. \end{equation*}

   Therefore, we require

   \begin{equation*} -a = 1, 2a-b = 0, b=c. \end{equation*}

   Hence, a particular solution is

   \begin{equation*} y = -t^2 - 2t - 2. \end{equation*}

   All together, we find that the general solution is

   \begin{equation*} y(t) = Ae^{t} -t^2 - 2t - 2. \end{equation*}

6. \(\ds 2y' +y=t\)

   Answer
   \(\ds y=Ae^{-t/2}+t-2\)
   Solution
   We first solve the corresponding homogeneous equation using separation of variables:

   \begin{equation*} \begin{split} 2\diff{y}{t} \amp=-y \\ \int \frac{1}{y}\,dy \amp= \int \frac{-1}{2} \,dt \\ \ln|y| \amp= -\frac{t}{2} + C\\ y(t) \amp= Ae^{-t/2}.\end{split} \end{equation*}

   We now use variation of parameters to find a particular solution. The right-hand side is linear, and so a particular solution must be of the form \(y = at+b.\) Now apply the DE:

   \begin{equation*} 2 \diff{}{t} (at+b) + (at+b) = at + 2a+b. \end{equation*}

   Therefore, a particular solution must satisfy

   \begin{equation*} a = 1, 2a+b = 0 \implies a = 1, b = -2. \end{equation*}

   All together, we find that the general solution is

   \begin{equation*} y(t) = Ae^{-t/2} + t-2. \end{equation*}

Find the general solution of the following non-homogeneous differential equations on the restricted domain.

1. \(\ds ty' -2y=1/t\text{,}\) \(t>0\)

   Answer
   \(\ds y=At^2-{1\over3t}\)
   Solution
   We first solve the corresponding homogeneous equation:

   \begin{equation*} \begin{split} t\diff{y}{t} \amp= 2y\\ \int \frac{1}{y} \,dy \amp= \int \frac{2}{t} \,dt \\ \ln|y| \amp= 2\ln|t| + C\\ y \amp= Ae^{\ln(t^2)}\\ \amp= At^2. \end{split} \end{equation*}

   We now look for a particular solution on the restricted domain. We look for a solution of the form \(b/t + c\text{:}\)

   \begin{equation*} t\left(- \frac{b}{t^2}\right) - 2\left(\frac{b}{t} + c\right) = \frac{-3b}{t} + 2c. \end{equation*}

   Therefore, we need \(b=-1/3\) and \(c=0\text{.}\) Hence, the general solution is

   \begin{equation*} y(t) = At^2 -\frac{1}{3t}. \end{equation*}

2. \(\ds ty'+y=\sqrt{t}\text{,}\) \(t>0\)

   Answer
   \(\ds y={c\over t}+{2\over3}\sqrt t\)
   Solution
   We first solve the corresponding homogeneous equation:

   \begin{equation*} \begin{split} t\diff{y}{t} \amp= -y\\ \int \frac{1}{y} \,dy \amp= -\int \frac{2}{t} \,dt \\ \ln|y| \amp= -\ln|t| + C\\ y \amp= Ae^{\ln(t)^{-1}}\\ \amp= \frac{A}{t}. \end{split} \end{equation*}

   We now look for a particular solution on the restricted domain. We look for a solution of the form \(a\sqrt{t} + c\text{:}\)

   \begin{equation*} t\left(\frac{a}{2\sqrt{t}}\right) + \left(a\sqrt{t} + c\right) = \left(\frac{a}{2} + a\right)\sqrt{t} + c. \end{equation*}

   Therefore, we need \(a=2/3\) and \(c=0\text{.}\) Hence, the general solution is

   \begin{equation*} y(t) = \frac{A}{t} + \frac{2}{3}\sqrt{t}. \end{equation*}

3. \(\ds y'\cos t+y\sin t=1\text{,}\) \(-\pi/2\lt t\lt \pi/2\)

   Answer
   \(\ds y= A\cos t+\sin t\)
   Solution
   We first solve the corresponding homogeneous equation:

   \begin{equation*} \begin{split} \cos(t)\diff{y}{t}\amp= -y\sin(t)\\ \int \frac{1}{y} \,dy \amp= -\int \frac{\sin (t)}{\cos(t)} \,dt \\ \ln|y| \amp= -\int \tan(t) \,dt \end{split} \end{equation*}

   On this restricted domain, we know that

   \begin{equation*} \int \tan(t) \,dt = \ln|\cos(t)|\,dt + C. \end{equation*}

   Therefore, the general solution is

   \begin{equation*} y = A\cos(t). \end{equation*}

   To find a particular solution, we notice that when \(y=\sin(t)\text{,}\) we have

   \begin{equation*} (\sin(t)')\cos(t) + \sin^2(t) = \cos^2(t) + \sin^2(t) = 1. \end{equation*}

   Hence, this is a particular solution and so the general solution is

   \begin{equation*} y(t) = A\cos(t) +\sin(t). \end{equation*}

4. \(\ds y' + y\sec t=\tan t\text{,}\) \(-\pi/2\lt t\lt \pi/2\)

   Answer
   \(\ds y= {A\over\sec t+\tan t}+1-{t\over\sec t+\tan t}\)

Solve the following initial value problems concerning non-homogeneous DEs.

1. \(\ds y' +4y=8, y(0)=1\)

   Answer
   \(\ds y=2-e^{-4t}\)
   Solution
   We first solve the corresponding homogeneous equation using separation of variables:

   \begin{equation*} \begin{split} \diff{y}{t} \amp= -4y \\ \int \frac{1}{y}\,dy \amp= -4\int \,dt \\ \ln|y| \amp= -4t+ C\\ y(t) \amp= Ae^{-4t}.\end{split} \end{equation*}

   We now notice that \(y=2\) is a particular solution to the DE:

   \begin{equation*} \diff{}{t} (2) + 4(2) = 8. \end{equation*}

   Hence, the general solution is

   \begin{equation*} y(t) = Ae^{-4t} + 2. \end{equation*}

   We now apply the initial condition:

   \begin{equation*} y(0) = 1 \implies A + 2 = 1. \end{equation*}

   Therefore, the solution to the IVP is

   \begin{equation*} y(t) = -e^{-4t} + 2. \end{equation*}

2. \(\ds y'-2y=6,y(0)=3\)

   Answer
   \(\ds y=6e^{2t}-3\)
   Solution
   We first solve the corresponding homogeneous equation using separation of variables:

   \begin{equation*} \begin{split} \diff{y}{t} \amp= 2y \\ \int \frac{1}{y}\,dy \amp= 2\int \,dt \\ \ln|y| \amp= 2t+ C\\ y(t) \amp= Ae^{2t}.\end{split} \end{equation*}

   We now notice that \(y=-3\) is a particular solution to the DE:

   \begin{equation*} \diff{}{t} (-3) -2(-3) = 6. \end{equation*}

   Hence, the general solution is

   \begin{equation*} y(t) = Ae^{2t} -3.\ \end{equation*}

   We now apply the initial condition:

   \begin{equation*} y(0) = 3 \implies A -3 = 3. \end{equation*}

   Therefore, the solution to the IVP is

   \begin{equation*} y(t) =6e^{2t} -3. \end{equation*}

3. \(\ds y' +ty=5t, y(2)=1\)

   Answer
   \(\ds y=5-4e^{2-(1/2)t^2}\)
   Solution

   We first look for the general solution to the corresponding homogeneous problem,

   \begin{equation*} y' + ty = 0\text{.} \end{equation*}

   Separating variables, we find

   \begin{equation*} \begin{split} \diff{y}{t} \amp = -ty \\ \implies \int \frac{dy}{y} \amp = -\int t\,dt \\ \implies \ln |y| \amp = -\frac{1}{2}t^2+ C \\ \implies y \amp = A e^{-\frac{1}{2}t^2} \end{split} \end{equation*}

   So we take \(h(t)=A e^{-\frac{1}{2}t^2}\text{.}\) It remains to find any particular solution \(g(t)\text{:}\) we notice that \(g(t)=5\) is one such solution, since

   \begin{equation*} \diff{}{t} (5) + t \cdot 5 = 5t\text{.} \end{equation*}

   Therefore, the general solution is

   \begin{equation*} y(t) = h(t) + g(t) = A e^{-\frac{1}{2}t^2} + 5\text{.} \end{equation*}

   Now apply the initial condition:

   \begin{equation*} y(2) = 1 \implies Ae^{-2} + 5 = 1 \implies A = -4e^2\text{.} \end{equation*}

   Altogether, we find that

   \begin{equation*} y(t) = 5-4e^{2-\frac{1}{2}t^2}\text{.} \end{equation*}

4. \(\ds y'+e^ty=-2e^t, y(0)=e^{-1}\)

   Answer
   \(\ds y=e^{-e^t}-2\)

5. \(\ds y'-y=t^2, y(0)=4\)

   Answer
   \(\ds y=6e^{t}-t^2-2t-2\)
   Solution

   We use the method of integrating factors. Let

   \begin{equation*} I(t) = e^{\int -1 \,dt} = e^{-t}\text{,} \end{equation*}

   where we took the arbitrary constant of integration to be zero. Multiplying the DE by \(I(t)\) gives

   \begin{equation*} e^{-t}y' - e^{-t}y = e^{-t}t^2 \implies \diff{}{t}\left[e^{-t}y\right] = e^{-t}t^2\text{.} \end{equation*}

   Integrating, we find

   \begin{equation*} e^{-t}y = \int e^{-t}t^2\,dt\text{.} \end{equation*}

   We solve using integration by parts with

   \begin{equation*} \begin{array}{ll} u = t^2 \amp dv = e^{-t}\,dt \\ du = 2t\,dt \amp v = -e^{-t} \end{array} \end{equation*}

   So,

   \begin{equation*} \int t^2 e^{-t}\,dt = -t^2e^{-t} + 2\int t e^{-t}\,dt\text{,} \end{equation*}

   and apply integration by parts again to find

   \begin{equation*} \int t^2 e^{-t}\,dt = -e^{-t} \left(t^2+2t+2\right) + C\text{.} \end{equation*}

   Thus, we have

   \begin{equation*} e^{-t}y = -e^{-t}\left(t^2+2t+2\right) + C \implies y(t) = Ce^t - (t^2+2t+2)\text{.} \end{equation*}

   (Notice that the particular solution, \(g(t)=-t^2-2t-2\) may have been difficult to 鈥済uess鈥� .) We now apply the initial condition:

   \begin{equation*} y(0) = -2 + C = 4 \implies C = 6\text{.} \end{equation*}

   Therefore, the solution to the initial value problem is

   \begin{equation*} y(t) = 6e^t - (t^2+2t+2)\text{.} \end{equation*}

6. \(\ds 2y' +y=t, y(1)=-1\)

   Answer
   \(\ds y=t-2\)
   Solution
   We first solve the corresponding homogeneous equation using separation of variables:

   \begin{equation*} \begin{split} 2\diff{y}{t} \amp=-y \\ \int \frac{1}{y}\,dy \amp= \int \frac{-1}{2} \,dt \\ \ln|y| \amp= -\frac{t}{2} + C\\ y(t) \amp= Ae^{-t/2}.\end{split} \end{equation*}

   We now use variation of parameters to find a particular solution. The right-hand side is linear, and so a particular solution must be of the form \(y = at+b.\) Now apply the DE:

   \begin{equation*} 2 \diff{}{t} (at+b) + (at+b) = at + 2a+b. \end{equation*}

   Therefore, a particular solution must satisfy

   \begin{equation*} a = 1, 2a+b = 0 \implies a = 1, b = -2. \end{equation*}

   All together, we find that the general solution is

   \begin{equation*} y(t) = Ae^{-t/2} + t-2. \end{equation*}

   We now apply the initial condition:

   \begin{equation*} y(1) = -1 \implies Ae^{1/2} + 1 -2 = -1 \implies A= 0. \end{equation*}

   Therefore, the general solution is

   \begin{equation*} y(t) = t-2. \end{equation*}