The givin system of ODEs was y\'1 = y1+y2 adn y\'2 = 3y1-y2. I have a solution o

Question

The givin system of ODEs was y'1 = y1+y2 adn y'2 = 3y1-y2. I have a solution of: y1 =C1*2e^(2t)-2C2*e^(-2t) and y2 = -C1*e^(2t)+3C2*e^(-2t).

For the solution provided , write the phase plane solution by eliminating t from the solution and writing equaiton with y1 and y2.

NOTE:The final solution should represent an equaiton of some kind of shape, such as circle, hyperbola, elipse, etc......

Explanation / Answer

y1 =2*C1e^(2t)-2C2*e^(-2t) => y1*e^(2t) = 2*C1e^(4t) - 2C2 -------> Equation 1 y2 =-C1*e^(2t)+3C2*e^(-2t) => y2*e^(2t) = -C1e^(4t) + 3C2 -------> Equation 2. (y1 + 2*y2) * e^(2t) = 4C2 => e^(2t) = 4C2 / (y1 + 2*y2) (3y1 + 2y2)* e^(2t) = 4C1 * e^(4t) => e^(2t) = (3y1 + 2y2) / 4C1 hence , (3y1 + 2y2) / 4C1 = 4C2 / (y1 + 2*y2) =>(3y1 + 2y2) * (y1 + 2*y2) = 16*C1*C2 =>3y1^2 + 4y2^2 + 8y1*y2 = 16*C1*C2 which is of the form : ax^2 + b*x*y + cy^2 =K b^2 - 4ac = 64 - 4*3*4 = 16 >0. hence the above equation 3y1^2 + 4y2^2 + 8y1*y2 = 16*C1*C2 is Hyperbola.

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