# Writing Prompt(s) One method f

Writing Prompt(s)

One method for solving a system of first order lineardifferential equation such as

x ′ = a x + b y y ′ = c x + d y

is to take the derivative of the first equation and use thesecond equation to “decouple” the system and create a secondorder equation, which we can solve using our previous techniques.Does this always work? If not, what conditions on the constants a,b, c, and d must be enforced? If it does work, we can then arriveat an equation for x(t). How do we proceed in finding an equationfor y(t)?

Answer:

Using the method of elimination,the system of n lineardifferential equations reduces into a single nth order lineardifferential equation.This method is useful for simple cases likesystem of order 2.Now the given system isx’=ax+byy’=cx+dyWhere a, b, c, d are constant coefficients.In the method of elimination there is no such restriction onconstant coefficient a,b,c,d.But if you wish to find some nontrivial(non-constant solution) solution to the system one of themmust be non-zero.In method of elimination we do similar things as we do to findsolution to system of equation in linear algebra.In this method we first differentiate first equation and substitutey’ from the second equation and we get a second order lineardifferential in x which isx”-(a+d)x’+(ad-bc)=0Now this equation could be solve easily by finding the roots of thecharacteristic equationm^{2}-(a+d)m+(ad-bc)=0Let us suppose the general solution to x bex=C_{1}f_{1}+C_{2}f_{2} whereC_{1} and C_{2} are constants f_{1},f_{2} are parts of the solution and the function ofindependent variable only.Then by putting value of x in y’=cx+dy we gety’=c(C_{1}f_{1}+C_{2}f_{2})+dy=>y’-dy=c(C_{1}f_{1}+Cf_{2})Which is a first order first degree linear differential equationand finding the integrating factor I.F. and multiplying theequation with this I.F. ,then integrating we get the requiredsolution to y.(solving for y using integrating factor is one methodto solve, you could find other methods to as for yourcomfortable).