Find the solution to the given system of IVPs.
\[ 2 \, {x} = -9 \, {y} - {x'} \hspace{2em}x(0)= -8 \]
\[ -7 \, {y} - 4 \, {x} = {y'} \hspace{2em}x(0)= 5 \]
Answer:
\[x= -9 \, e^{\left(2 \, t\right)} + e^{\left(-11 \, t\right)} \]
\[y= 4 \, e^{\left(2 \, t\right)} + e^{\left(-11 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ 0 = -5 \, {x} + 2 \, {y} + {x'} \hspace{2em}x(0)= -3 \]
\[ -2 \, {x} = {y'} - 8 \, {y} \hspace{2em}x(0)= 1 \]
Answer:
\[x= -e^{\left(9 \, t\right)} - 2 \, e^{\left(4 \, t\right)} \]
\[y= 2 \, e^{\left(9 \, t\right)} - e^{\left(4 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -{x'} - 4 \, {y} + 5 \, {x} = 0 \hspace{2em}x(0)= -3 \]
\[ {x} = 8 \, {y} - {y'} \hspace{2em}x(0)= -2 \]
Answer:
\[x= e^{\left(9 \, t\right)} - 4 \, e^{\left(4 \, t\right)} \]
\[y= -e^{\left(9 \, t\right)} - e^{\left(4 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -2 \, {y} = -2 \, {x} + {x'} \hspace{2em}x(0)= 3 \]
\[ 0 = -2 \, {x} + 5 \, {y} - {y'} \hspace{2em}x(0)= -1 \]
Answer:
\[x= e^{\left(6 \, t\right)} + 2 \, e^{t} \]
\[y= -2 \, e^{\left(6 \, t\right)} + e^{t} \]
Find the solution to the given system of IVPs.
\[ {x'} - 2 \, {x} - 4 \, {y} = 0 \hspace{2em}x(0)= -5 \]
\[ 5 \, {y} = -{x} + {y'} \hspace{2em}x(0)= 0 \]
Answer:
\[x= -e^{\left(6 \, t\right)} - 4 \, e^{t} \]
\[y= -e^{\left(6 \, t\right)} + e^{t} \]
Find the solution to the given system of IVPs.
\[ 2 \, {y} - 4 \, {x} = -{x'} \hspace{2em}x(0)= 1 \]
\[ 7 \, {y} = 2 \, {x} + {y'} \hspace{2em}x(0)= 3 \]
Answer:
\[x= -e^{\left(8 \, t\right)} + 2 \, e^{\left(3 \, t\right)} \]
\[y= 2 \, e^{\left(8 \, t\right)} + e^{\left(3 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ {x'} = 3 \, {x} - 4 \, {y} \hspace{2em}x(0)= 5 \]
\[ 8 \, {y} - {y'} = -{x} \hspace{2em}x(0)= -2 \]
Answer:
\[x= e^{\left(7 \, t\right)} + 4 \, e^{\left(4 \, t\right)} \]
\[y= -e^{\left(7 \, t\right)} - e^{\left(4 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -{y} + {x} = -{x'} \hspace{2em}x(0)= 0 \]
\[ 0 = 6 \, {y} + 4 \, {x} + {y'} \hspace{2em}x(0)= -3 \]
Answer:
\[x= -e^{\left(-2 \, t\right)} + e^{\left(-5 \, t\right)} \]
\[y= e^{\left(-2 \, t\right)} - 4 \, e^{\left(-5 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -9 \, {y} = {x} - {x'} \hspace{2em}x(0)= 10 \]
\[ 4 \, {x} - {y'} - 4 \, {y} = 0 \hspace{2em}x(0)= 3 \]
Answer:
\[x= 9 \, e^{\left(5 \, t\right)} + e^{\left(-8 \, t\right)} \]
\[y= 4 \, e^{\left(5 \, t\right)} - e^{\left(-8 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -4 \, {y} = -2 \, {x} + {x'} \hspace{2em}x(0)= 5 \]
\[ {x} = 5 \, {y} - {y'} \hspace{2em}x(0)= 0 \]
Answer:
\[x= e^{\left(6 \, t\right)} + 4 \, e^{t} \]
\[y= -e^{\left(6 \, t\right)} + e^{t} \]
Find the solution to the given system of IVPs.
\[ {x'} + 2 \, {x} + {y} = 0 \hspace{2em}x(0)= 0 \]
\[ -7 \, {y} = -4 \, {x} + {y'} \hspace{2em}x(0)= 3 \]
Answer:
\[x= -e^{\left(-3 \, t\right)} + e^{\left(-6 \, t\right)} \]
\[y= -e^{\left(-3 \, t\right)} + 4 \, e^{\left(-6 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ {x'} = -2 \, {x} + {y} \hspace{2em}x(0)= 0 \]
\[ -4 \, {x} + 5 \, {y} = -{y'} \hspace{2em}x(0)= -5 \]
Answer:
\[x= -e^{\left(-t\right)} + e^{\left(-6 \, t\right)} \]
\[y= -e^{\left(-t\right)} - 4 \, e^{\left(-6 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ {x'} = {y} - {x} \hspace{2em}x(0)= 0 \]
\[ 6 \, {y} + {y'} + 4 \, {x} = 0 \hspace{2em}x(0)= 3 \]
Answer:
\[x= e^{\left(-2 \, t\right)} - e^{\left(-5 \, t\right)} \]
\[y= -e^{\left(-2 \, t\right)} + 4 \, e^{\left(-5 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ {x'} - {y} = 2 \, {x} \hspace{2em}x(0)= -2 \]
\[ {y'} + 3 \, {y} + 4 \, {x} = 0 \hspace{2em}x(0)= 5 \]
Answer:
\[x= -e^{\left(-2 \, t\right)} - e^{t} \]
\[y= 4 \, e^{\left(-2 \, t\right)} + e^{t} \]
Find the solution to the given system of IVPs.
\[ 0 = -4 \, {x} + {x'} + 4 \, {y} \hspace{2em}x(0)= 5 \]
\[ 9 \, {y} = -{x} + {y'} \hspace{2em}x(0)= -2 \]
Answer:
\[x= e^{\left(8 \, t\right)} + 4 \, e^{\left(5 \, t\right)} \]
\[y= -e^{\left(8 \, t\right)} - e^{\left(5 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -{x'} = -3 \, {y} + 4 \, {x} \hspace{2em}x(0)= -4 \]
\[ {y'} - {y} = 12 \, {x} \hspace{2em}x(0)= 1 \]
Answer:
\[x= -e^{\left(5 \, t\right)} - 3 \, e^{\left(-8 \, t\right)} \]
\[y= -3 \, e^{\left(5 \, t\right)} + 4 \, e^{\left(-8 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ {x'} = {x} + 9 \, {y} \hspace{2em}x(0)= -8 \]
\[ 0 = 4 \, {x} - {y'} - 4 \, {y} \hspace{2em}x(0)= -5 \]
Answer:
\[x= -9 \, e^{\left(5 \, t\right)} + e^{\left(-8 \, t\right)} \]
\[y= -4 \, e^{\left(5 \, t\right)} - e^{\left(-8 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -{x} + {y} - {x'} = 0 \hspace{2em}x(0)= -2 \]
\[ -2 \, {y} + {y'} = 4 \, {x} \hspace{2em}x(0)= -3 \]
Answer:
\[x= -e^{\left(3 \, t\right)} - e^{\left(-2 \, t\right)} \]
\[y= -4 \, e^{\left(3 \, t\right)} + e^{\left(-2 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -9 \, {y} = {x'} + 3 \, {x} \hspace{2em}x(0)= 10 \]
\[ -{y'} - 4 \, {x} = 8 \, {y} \hspace{2em}x(0)= -3 \]
Answer:
\[x= e^{\left(-12 \, t\right)} + 9 \, e^{t} \]
\[y= e^{\left(-12 \, t\right)} - 4 \, e^{t} \]
Find the solution to the given system of IVPs.
\[ -{y} = -{x'} - 2 \, {x} \hspace{2em}x(0)= 2 \]
\[ -4 \, {x} = -{y'} - 5 \, {y} \hspace{2em}x(0)= -3 \]
Answer:
\[x= e^{\left(-t\right)} + e^{\left(-6 \, t\right)} \]
\[y= e^{\left(-t\right)} - 4 \, e^{\left(-6 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -4 \, {x} + {y} + {x'} = 0 \hspace{2em}x(0)= 0 \]
\[ -4 \, {x} - {y'} = -7 \, {y} \hspace{2em}x(0)= 5 \]
Answer:
\[x= -e^{\left(8 \, t\right)} + e^{\left(3 \, t\right)} \]
\[y= 4 \, e^{\left(8 \, t\right)} + e^{\left(3 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ 3 \, {y} - 3 \, {x} = {x'} \hspace{2em}x(0)= -4 \]
\[ 8 \, {y} + {y'} - 12 \, {x} = 0 \hspace{2em}x(0)= -1 \]
Answer:
\[x= -e^{\left(-12 \, t\right)} - 3 \, e^{t} \]
\[y= 3 \, e^{\left(-12 \, t\right)} - 4 \, e^{t} \]
Find the solution to the given system of IVPs.
\[ 2 \, {x} - {x'} = 3 \, {y} \hspace{2em}x(0)= -4 \]
\[ 7 \, {y} = 12 \, {x} + {y'} \hspace{2em}x(0)= -1 \]
Answer:
\[x= -e^{\left(11 \, t\right)} - 3 \, e^{\left(-2 \, t\right)} \]
\[y= 3 \, e^{\left(11 \, t\right)} - 4 \, e^{\left(-2 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ {y} = 3 \, {x} + {x'} \hspace{2em}x(0)= 2 \]
\[ 2 \, {y} = 4 \, {x} + {y'} \hspace{2em}x(0)= 5 \]
Answer:
\[x= e^{\left(-2 \, t\right)} + e^{t} \]
\[y= e^{\left(-2 \, t\right)} + 4 \, e^{t} \]
Find the solution to the given system of IVPs.
\[ 0 = -{x} - 2 \, {y} + {x'} \hspace{2em}x(0)= 1 \]
\[ 2 \, {y} = -{y'} + 2 \, {x} \hspace{2em}x(0)= 3 \]
Answer:
\[x= 2 \, e^{\left(2 \, t\right)} - e^{\left(-3 \, t\right)} \]
\[y= e^{\left(2 \, t\right)} + 2 \, e^{\left(-3 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -4 \, {x} = -{x'} - 12 \, {y} \hspace{2em}x(0)= 1 \]
\[ {y} = -{y'} - 3 \, {x} \hspace{2em}x(0)= 4 \]
Answer:
\[x= -3 \, e^{\left(8 \, t\right)} + 4 \, e^{\left(-5 \, t\right)} \]
\[y= e^{\left(8 \, t\right)} + 3 \, e^{\left(-5 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ 0 = {x} - {y} - {x'} \hspace{2em}x(0)= -2 \]
\[ {y'} = -4 \, {x} - 2 \, {y} \hspace{2em}x(0)= -3 \]
Answer:
\[x= -e^{\left(2 \, t\right)} - e^{\left(-3 \, t\right)} \]
\[y= e^{\left(2 \, t\right)} - 4 \, e^{\left(-3 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -{x'} - {x} + 2 \, {y} = 0 \hspace{2em}x(0)= -3 \]
\[ {y'} = -2 \, {x} - 6 \, {y} \hspace{2em}x(0)= 3 \]
Answer:
\[x= -2 \, e^{\left(-2 \, t\right)} - e^{\left(-5 \, t\right)} \]
\[y= e^{\left(-2 \, t\right)} + 2 \, e^{\left(-5 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ 0 = -4 \, {x} + {x'} + 4 \, {y} \hspace{2em}x(0)= 5 \]
\[ 9 \, {y} = -{x} + {y'} \hspace{2em}x(0)= -2 \]
Answer:
\[x= e^{\left(8 \, t\right)} + 4 \, e^{\left(5 \, t\right)} \]
\[y= -e^{\left(8 \, t\right)} - e^{\left(5 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ {y} - 5 \, {x} = -{x'} \hspace{2em}x(0)= 0 \]
\[ -2 \, {y} + {y'} = -4 \, {x} \hspace{2em}x(0)= 5 \]
Answer:
\[x= -e^{\left(6 \, t\right)} + e^{t} \]
\[y= e^{\left(6 \, t\right)} + 4 \, e^{t} \]
Find the solution to the given system of IVPs.
\[ {x} + 4 \, {y} = {x'} \hspace{2em}x(0)= -3 \]
\[ {x} = {y'} + 2 \, {y} \hspace{2em}x(0)= -2 \]
Answer:
\[x= -4 \, e^{\left(2 \, t\right)} + e^{\left(-3 \, t\right)} \]
\[y= -e^{\left(2 \, t\right)} - e^{\left(-3 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ 0 = -9 \, {y} - {x'} + {x} \hspace{2em}x(0)= -8 \]
\[ -6 \, {y} + 4 \, {x} = -{y'} \hspace{2em}x(0)= -5 \]
Answer:
\[x= e^{\left(10 \, t\right)} - 9 \, e^{\left(-3 \, t\right)} \]
\[y= -e^{\left(10 \, t\right)} - 4 \, e^{\left(-3 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -{x'} = -4 \, {y} + 2 \, {x} \hspace{2em}x(0)= -3 \]
\[ -3 \, {y} + {x} = -{y'} \hspace{2em}x(0)= 0 \]
Answer:
\[x= e^{\left(2 \, t\right)} - 4 \, e^{\left(-t\right)} \]
\[y= e^{\left(2 \, t\right)} - e^{\left(-t\right)} \]
Find the solution to the given system of IVPs.
\[ -4 \, {y} - {x'} - 2 \, {x} = 0 \hspace{2em}x(0)= 5 \]
\[ {x} = -{y'} - 5 \, {y} \hspace{2em}x(0)= 0 \]
Answer:
\[x= 4 \, e^{\left(-t\right)} + e^{\left(-6 \, t\right)} \]
\[y= -e^{\left(-t\right)} + e^{\left(-6 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ 0 = {x'} + 2 \, {y} + 2 \, {x} \hspace{2em}x(0)= -1 \]
\[ 5 \, {y} + 2 \, {x} + {y'} = 0 \hspace{2em}x(0)= 3 \]
Answer:
\[x= -2 \, e^{\left(-t\right)} + e^{\left(-6 \, t\right)} \]
\[y= e^{\left(-t\right)} + 2 \, e^{\left(-6 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -{x'} = -{y} - 5 \, {x} \hspace{2em}x(0)= -2 \]
\[ -{y'} = -8 \, {y} - 4 \, {x} \hspace{2em}x(0)= -3 \]
Answer:
\[x= -e^{\left(9 \, t\right)} - e^{\left(4 \, t\right)} \]
\[y= -4 \, e^{\left(9 \, t\right)} + e^{\left(4 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -{x'} - 2 \, {x} + {y} = 0 \hspace{2em}x(0)= -2 \]
\[ 4 \, {x} = {y'} + 5 \, {y} \hspace{2em}x(0)= 3 \]
Answer:
\[x= -e^{\left(-t\right)} - e^{\left(-6 \, t\right)} \]
\[y= -e^{\left(-t\right)} + 4 \, e^{\left(-6 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ 6 \, {y} = -{x} - {x'} \hspace{2em}x(0)= -5 \]
\[ -6 \, {x} = {y'} + 6 \, {y} \hspace{2em}x(0)= -1 \]
Answer:
\[x= -3 \, e^{\left(3 \, t\right)} - 2 \, e^{\left(-10 \, t\right)} \]
\[y= 2 \, e^{\left(3 \, t\right)} - 3 \, e^{\left(-10 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ 2 \, {y} = {x'} - 4 \, {x} \hspace{2em}x(0)= 1 \]
\[ -{y'} + 2 \, {x} = -7 \, {y} \hspace{2em}x(0)= -3 \]
Answer:
\[x= -e^{\left(8 \, t\right)} + 2 \, e^{\left(3 \, t\right)} \]
\[y= -2 \, e^{\left(8 \, t\right)} - e^{\left(3 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -4 \, {x} + {y} = -{x'} \hspace{2em}x(0)= -2 \]
\[ -7 \, {y} + {y'} = -4 \, {x} \hspace{2em}x(0)= 3 \]
Answer:
\[x= -e^{\left(8 \, t\right)} - e^{\left(3 \, t\right)} \]
\[y= 4 \, e^{\left(8 \, t\right)} - e^{\left(3 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -{x'} = -{x} - 12 \, {y} \hspace{2em}x(0)= -1 \]
\[ 3 \, {x} = 4 \, {y} + {y'} \hspace{2em}x(0)= 4 \]
Answer:
\[x= 3 \, e^{\left(5 \, t\right)} - 4 \, e^{\left(-8 \, t\right)} \]
\[y= e^{\left(5 \, t\right)} + 3 \, e^{\left(-8 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ 2 \, {y} = -5 \, {x} + {x'} \hspace{2em}x(0)= -1 \]
\[ 8 \, {y} + 2 \, {x} - {y'} = 0 \hspace{2em}x(0)= 3 \]
Answer:
\[x= e^{\left(9 \, t\right)} - 2 \, e^{\left(4 \, t\right)} \]
\[y= 2 \, e^{\left(9 \, t\right)} + e^{\left(4 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ 2 \, {y} + 3 \, {x} = {x'} \hspace{2em}x(0)= 1 \]
\[ -8 \, {y} + 2 \, {x} = -{y'} \hspace{2em}x(0)= -1 \]
Answer:
\[x= -e^{\left(7 \, t\right)} + 2 \, e^{\left(4 \, t\right)} \]
\[y= -2 \, e^{\left(7 \, t\right)} + e^{\left(4 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -{y} + {x'} = -2 \, {x} \hspace{2em}x(0)= -2 \]
\[ 0 = {y} + 4 \, {x} - {y'} \hspace{2em}x(0)= -3 \]
Answer:
\[x= -e^{\left(2 \, t\right)} - e^{\left(-3 \, t\right)} \]
\[y= -4 \, e^{\left(2 \, t\right)} + e^{\left(-3 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ {y} - 2 \, {x} = {x'} \hspace{2em}x(0)= 0 \]
\[ {y} + 4 \, {x} = {y'} \hspace{2em}x(0)= 5 \]
Answer:
\[x= e^{\left(2 \, t\right)} - e^{\left(-3 \, t\right)} \]
\[y= 4 \, e^{\left(2 \, t\right)} + e^{\left(-3 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ 2 \, {y} = {x'} - 4 \, {x} \hspace{2em}x(0)= 1 \]
\[ 2 \, {x} = -7 \, {y} + {y'} \hspace{2em}x(0)= -3 \]
Answer:
\[x= -e^{\left(8 \, t\right)} + 2 \, e^{\left(3 \, t\right)} \]
\[y= -2 \, e^{\left(8 \, t\right)} - e^{\left(3 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ {y} - 4 \, {x} - {x'} = 0 \hspace{2em}x(0)= 0 \]
\[ -9 \, {y} - {y'} = 4 \, {x} \hspace{2em}x(0)= 3 \]
Answer:
\[x= e^{\left(-5 \, t\right)} - e^{\left(-8 \, t\right)} \]
\[y= -e^{\left(-5 \, t\right)} + 4 \, e^{\left(-8 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ 2 \, {y} - {x'} = -4 \, {x} \hspace{2em}x(0)= 1 \]
\[ -{y'} = -9 \, {y} + 2 \, {x} \hspace{2em}x(0)= -1 \]
Answer:
\[x= -e^{\left(8 \, t\right)} + 2 \, e^{\left(5 \, t\right)} \]
\[y= -2 \, e^{\left(8 \, t\right)} + e^{\left(5 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ -{x'} - {x} = 4 \, {y} \hspace{2em}x(0)= 5 \]
\[ 0 = 2 \, {y} - {x} - {y'} \hspace{2em}x(0)= 0 \]
Answer:
\[x= e^{\left(3 \, t\right)} + 4 \, e^{\left(-2 \, t\right)} \]
\[y= -e^{\left(3 \, t\right)} + e^{\left(-2 \, t\right)} \]
Find the solution to the given system of IVPs.
\[ {x} - {y} = {x'} \hspace{2em}x(0)= -2 \]
\[ -4 \, {x} - 2 \, {y} - {y'} = 0 \hspace{2em}x(0)= -3 \]
Answer:
\[x= -e^{\left(2 \, t\right)} - e^{\left(-3 \, t\right)} \]
\[y= e^{\left(2 \, t\right)} - 4 \, e^{\left(-3 \, t\right)} \]