The question is
For the circuit shown in Figure, find the currents using Laplace transformation and without using Laplace transformation assuming zero currents and charge when $t=0.$
The equations written using Kirchhoff's Voltage Law (mesh analysis) are \begin{align*} I'_1 =&-I_1R_1/L+I_2 R_1/L + E/L &\\ I'_2 =&-I_2/(C(R_1+R_2))+(-I_1R_1/L+I_2 R_1/L + E/L)R_1/(R_1+R_2) \end{align*}
The first part was done using
$\begin{alignat}{1} \left[ \begin{array}{r} I'_1\\I'_2 \end{array} \right] = &\left[ \begin{array}{rr} -R_1/L & R_1/L \\ -R^2_1/(L(R_1+R_2)) &-1/(C(R_1+R_2))+( R^2_1/(L(R_1+R_2) ) \end{array} \right] \left[ \begin{array}{r} I_1\\I_2 \end{array} \right] \\+ &\left[ \begin{array}{r} E/L\\ER_1/(L(R_1+R_2)) \end{array} \right] \end{alignat} $
and the answer is
$\mathbf{y}=139.44\mathsf{e}^{-1.54 t}\left[ \begin{array}{rr} 2 \\ 0.46 \end{array} \right] -47.58 \mathsf{e}^{-0.26 t}\left[ \begin{array}{rr} 2 \\ 1.74 \end{array} \right] +\left[ \begin{array}{rr} 100 \\ 0 \end{array} \right]$
Now, for the second part using Laplace transformation, I'm not getting the correct answer.
$\left[ \begin{array}{rr} s+R_1/L & -R_1/L \\ R^2_1/(L(R_1+R_2)) &s+1/(C(R_1+R_2))-( R^2_1/(L(R_1+R_2) ) \end{array} \right]\left[ \begin{array}{r} I_1(s)\\I_2(s) \end{array} \right] = \left[ \begin{array}{r} E/sL\\ER_1/(sL(R_1+R_2)) \end{array} \right]$
For instance, for $i_2(t),$ I'm getting
$\displaystyle \frac{400\,\sqrt{41}\,{\mathrm{e}}^{-\frac{9\,t}{10}} \,\sinh \left(\frac{\sqrt{41}\,t}{10}\right)}{41}$
Why am I not getting the same answer when using Laplace Transform?
Note that: $R_1=2\,\Omega$, $R_2=8\,\Omega$, $L=1\,\text{H}$, $C=0.5\,\text{F}$, $E=200\,\text{V}$
Update: Thats how I solved the problem as a system of differential equations
$ y_1=\mathsf{e}^{ \big(-\frac{9}{10}\big)t} \Bigg( -\frac{200\,\sqrt{41}}{41} \mathsf{e}^{ \big(-\frac{\sqrt{41}}{10}\big)t}\big( \frac{\sqrt{41}}{4}+\frac{11}{4}\big)+ \frac{200\,\sqrt{41}}{41}\mathsf{e}^{\big(\frac{\sqrt{41}}{10}\big)t}\big( -\frac{\sqrt{41}}{4}+\frac{11}{4} \big)\Bigg) + 100 \\ =-\frac{200\,\sqrt{41}}{41}\mathsf{e}^{ \big(-\frac{9}{10}\big)t} \Bigg( \mathsf{e}^{ \big(-\frac{\sqrt{41}}{10}\big)t}\big( \frac{\sqrt{41}}{4}+\frac{11}{4}\big)+\mathsf{e}^{\big(\frac{\sqrt{41}}{10}\big)t}\big( \frac{\sqrt{41}}{4}-\frac{11}{4} \big) \Bigg) + 100 \\ =-\frac{200\,\sqrt{41}}{41}\mathsf{e}^{ \big(-\frac{9}{10}\big)t} \Bigg( \frac{\sqrt{41}}{4}\Bigg(\mathsf{e}^{ \big(-\frac{\sqrt{41}}{10}\big)t}+\mathsf{e}^{ \big(-\frac{\sqrt{41}}{10}\big)t}\Bigg) -\frac{11}{4} \Bigg(\mathsf{e}^{\big(\frac{-\sqrt{41}}{10}\big)t} -\mathsf{e}^{\big(\frac{\sqrt{41}}{10}\big)t} \Bigg) + 100 \\ =-\frac{200\,\sqrt{41}}{41}\mathsf{e}^{ \big(-\frac{9}{10}\big)t} \Bigg( -\frac{\sqrt{41}}{2}\cosh\frac{\sqrt{41}}{10}t +\frac{11}{2} \sinh\frac{\sqrt{41}}{10}t \Bigg) + 100 \\ $
I'm not getting the correct polarity, what am I doing wrong here
Using the Differential Equation
$$ \begin{bmatrix} I'_1 \\ I'_2 \end{bmatrix} = \begin{bmatrix} -2 & 2 \\ -2/5 & 1/5 \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \end{bmatrix} +\begin{bmatrix} 200 \\ 40 \end{bmatrix} $$ Your time domain mesh analysis equations are correct, but the system solution is not. Using Wolfram Mathematica$^{(*)}$, I got:
$$ \begin{alignat}{1} i_1(t) &= \frac{100}{41} e^{^{-\frac{9t}{10}}}\! \cdot \! \left[11 \sqrt{41} \sinh \left(\frac{\sqrt{41} t}{10}\right)-41 \cosh \left(\frac{\sqrt{41} t}{10}\right)\right]+100 \\[5px] i_2(t) &= \frac{400 \sqrt{41} ~ e^{-\frac{9 t}{10}} \sinh \left(\frac{\sqrt{41} t}{10}\right)}{41} \end{alignat} $$
In terms of numerical values:
$$ \begin{alignat}{1} i_1(t) &\approx -135.90 e^{-1.54t} + 35.90 e^{-0.26t} \\[5px] i_2(t) &\approx -31.23 e^{-1.54t} + 31.23 e^{-0.26t} \end{alignat} $$
Let us solve it manually, step by step, so that you can identify your specific mistake(s).
1. The system and its solution
• General Solution = Homogeneous Solution + Particular Solution: $$ \vec{i}~\!'=A\cdot \vec i + \vec b \implies \vec i = \vec i_h + \vec i_p $$
• True (Actual) Solution = General Solution with the Initial Conditions applied: $$ \vec{i} ~~~ \text{and} ~~~ \vec i_0 = \vec{i}(0) = \vec0 $$ 2. The homogeneous solution
• Eigenvalues $\lambda_1$ and $\lambda_2$ of $A$: $$ \begin{align} \det \left( {A - \lambda I} \right) &= \begin{vmatrix} -2 - \lambda & 2 \\ -2/5 & 1/5 - \lambda \end{vmatrix} \\&= \frac 1 5 \cdot (5\lambda^2 + 9\lambda + 2) \quad \implies \lambda_1 = \frac{-9-\sqrt{41}}{10},~~\lambda_2 = \frac{-9+\sqrt{41}}{10} \end{align} $$
• Eigenvectors $\vec v_1$ and $\vec v_2$ of $A$: $$ \lambda_1 = \frac{-9-\sqrt{41}}{10}: \begin{bmatrix} -2 - \lambda_1 & \!\!\!2 \\ -2/5 & \!\!\!1/5 -\lambda_1 \end{bmatrix} \begin{bmatrix} p \\ q \end{bmatrix} = \begin{bmatrix} 0\\0 \end{bmatrix} \implies \vec v_1 = \begin{bmatrix} \left(11+\sqrt{41}\right)/4 \\ 1 \end{bmatrix} \\[25px] \lambda_2 = \frac{-9+\sqrt{41}}{10}: \begin{bmatrix} -2 - \lambda_2 & \!\!\!2 \\ -2/5 & \!\!\!1/5 -\lambda_2 \end{bmatrix} \begin{bmatrix} r \\ s \end{bmatrix} = \begin{bmatrix} 0\\0 \end{bmatrix} \implies \vec v_2 = \begin{bmatrix} \left(11-\sqrt{41}\right)/4 \\ 1 \end{bmatrix} $$ • Homogeneous solution: $$ \vec i_h = c_1e^{\lambda_1 t} \vec v_1 + c_2e^{\lambda_2 t} \vec v_2 = c_1e^{^{\frac{-9-\sqrt{41}}{10}t}} \begin{bmatrix} \left(11+\sqrt{41}\right)/4 \\ 1 \end{bmatrix} + c_2e^{^{\frac{-9+\sqrt{41}}{10}t}} \begin{bmatrix} \left(11-\sqrt{41}\right)/4 \\ 1 \end{bmatrix} $$
3. The particular solution
• Guess (linear polynomial in this case): $$ \vec i_p = \vec c ~t + \vec d = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}t +\begin{bmatrix} d_1 \\ d_2 \end{bmatrix} \implies \vec{i}_p~\!\!\!' = \vec c $$ • Plugging into the system: $$ \vec c = A\cdot \big(\vec c ~t + \vec d \big) + \vec b \implies \vec0 = \left(A \vec c\right)t + \big(A\vec d + \vec b - \vec c \big) \implies \begin{cases} \!\!A \vec c = \vec0 \\[1ex] \!\!A\vec d + \vec b - \vec c = \vec0 \end{cases} \\[35px] \begin{alignat}{1} A \vec c = \vec0 &\implies \begin{bmatrix} -2 & 2 \\ -2/5 & 1/5 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies \vec c = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\[5px] A \vec d = \vec c - \vec b &\implies \begin{bmatrix} -2 & 2 \\ -2/5 & 1/5 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} = \begin{bmatrix} 0-200 \\ 0-40 \end{bmatrix} \implies \vec d = \begin{bmatrix} 100 \\ 0 \end{bmatrix} \end{alignat} \\[2px] $$
• Particular solution: $$ \vec i_p = \vec c ~t + \vec d = \vec d = \begin{bmatrix} 100 \\ 0 \end{bmatrix} \\ $$
4. The general solution $$ \vec i = \vec i_h + \vec i_p = c_1e^{^{\frac{-9-\sqrt{41}}{10}t}} \begin{bmatrix} \left(11+\sqrt{41}\right)/4 \\ 1 \end{bmatrix} + c_2e^{^{\frac{-9+\sqrt{41}}{10}t}} \begin{bmatrix} \left(11-\sqrt{41}\right)/4 \\ 1 \end{bmatrix} + \begin{bmatrix} 100 \\ 0 \end{bmatrix} $$
5. The true solution
• Applying the initial conditions: $$ \begin{alignat}{1} \vec i_0 = \vec0 &\implies c_1 \begin{bmatrix} \left(11+\sqrt{41}\right)/4 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} \left(11-\sqrt{41}\right)/4 \\ 1 \end{bmatrix} + \begin{bmatrix} 100 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\[5px] &\implies c_1 = -\frac{200}{\sqrt{41}}, ~~ c_2 = \frac{200}{\sqrt{41}} \end{alignat} $$
• True solution: $$ \begin{alignat}{1} \vec i &= -\frac{200}{\sqrt{41}} \cdot e^{^{\frac{-9-\sqrt{41}}{10}t}} \begin{bmatrix} \left(11+\sqrt{41}\right)/4 \\ 1 \end{bmatrix} + \frac{200}{\sqrt{41}} \cdot e^{^{\frac{-9+\sqrt{41}}{10}t}} \begin{bmatrix} \left(11-\sqrt{41}\right)/4 \\ 1 \end{bmatrix} + \begin{bmatrix} 100 \\ 0 \end{bmatrix} \end{alignat} $$
• True solution (numerical values):
(Note that we obtained the same solution given by Mathematica for $i_1$ and $i_2$) $$ \bbox[6px,border:1.5px solid black] { \vec i(t) \approx -31.23 \cdot e^{-1.54t} \begin{bmatrix} 4.35 \\ 1 \end{bmatrix} + 31.23 \cdot e^{-0.26t} \begin{bmatrix} 1.15 \\ 1 \end{bmatrix} + \begin{bmatrix} 100 \\ 0 \end{bmatrix} } \\ $$
Using Laplace Transform
$$ \begin{bmatrix} s+2 & -2 \\ 2/5 & s-1/5 \end{bmatrix} \begin{bmatrix} I_1(s) \\ I_2(s) \end{bmatrix} = \begin{bmatrix} 200/s \\ 40/s \end{bmatrix} $$
Your Laplace domain matrix system is correct too. Let us solve it:
$$ \begin{bmatrix} I_1(s) \\ I_2(s) \end{bmatrix} = \begin{bmatrix} s+2 & -2 \\ 2/5 & s-1/5 \end{bmatrix}^{-1} \begin{bmatrix} 200/s \\ 40/s \end{bmatrix} \implies \begin{cases} I_1(s) = \dfrac{1000s+200}{5s^3+9s^2+2s} \\[1ex] I_2(s) = \dfrac{200}{5s^2+9s+2} \end{cases} $$
Now, anti-transforming, we get the same results:
$$ \begin{alignat}{1} i_1(t) &= \mathscr{L}^{-1}\left[ I_1(s) \right] ~&= \frac{100}{41} e^{^{-\frac{9t}{10}}}\! \cdot \! \left[11 \sqrt{41} \sinh \left(\frac{\sqrt{41} t}{10}\right)-41 \cosh \left(\frac{\sqrt{41} t}{10}\right)\right]+100 \\[5px] i_2(t) &= \mathscr{L}^{-1}\left[ I_2(s) \right] ~&= \frac{400 \sqrt{41} ~ e^{-\frac{9 t}{10}} \sinh \left(\frac{\sqrt{41} t}{10}\right)}{41} \end{alignat} $$
${(*)}$ Here is a screenshot of the commands used in Mathematica 12.0 and their results. Note that solution1 and solution2 are the same, the difference is that one is written with the exponential function and the other with hyperbolic functions, namely: $$ \begin{array}{c|c} \sinh(x) = \dfrac{e^x-e^{-x}}{2} \quad & \quad \cosh(x)=\dfrac{e^x+e^{-x}}{2} \end{array} $$