Find an O D E y^''+a y^'+b y=0 for the given basis. e^2 x,...

Key Concepts

Characteristic Equation

The characteristic equation is an algebraic equation derived from a linear homogeneous differential equation with constant coefficients by assuming solutions of the form e^(rx). The roots of this polynomial equation determine the form of the general solution. Each distinct real root corresponds to an independent exponential solution, making the process of finding the fundamental set systematic and predictable.

Linear Homogeneous Differential Equations

A linear homogeneous differential equation with constant coefficients is an equation of the form ay'' + by' + c y = 0, where the coefficients a, b, and c are constants and the equation is set to zero. The term 'homogeneous' indicates that there is no non-zero term independent of the function y. These equations are central in many fields because their solutions can often be superposed to obtain a general solution.

Fundamental Set of Solutions

A fundamental set (or basis) of solutions refers to a set of linearly independent solutions of a differential equation, such that any solution of the equation can be written as a linear combination of these solutions. The existence of a fundamental set is guaranteed by the theory of linear differential equations, and knowing this set allows one to describe the complete solution space of the differential equation.

Transcript

00:01 Hello everyone so according to this given problem we have two bases of given differential equation like this right so we can write the solution of differential equation in terms of linear combination of their basis right so let's consider the solution of the differential equation be y c1 the first basis here is to power 2x and the other basis c2 is to power x and the other basis c2 is to power x right now what we can do we will take the first derivative because it is required according to the problem so this will give you two times 7 erase to power 2x plus c2 erase to power x now we will calculate it for double prime so this will become simply 4 c1 erase to power 2x plus and this will give you c2 erase to power x as it is right so what we can do we can just put these values of y double prime y prime and y in the above equation so let's put the values there so we will get 4 c1 e raise to power 2x plus c2 erase to power x and for this a a multiplied with y prime we want to get 2c1 a erase to power 2x and a c2 e rise to power x right and for the other part we have b c1 eras to power 2x plus b c2 e raise to power x is equals to 0 so what we need to do we need to find the values of a and b so let's consider erase to our 2x components first so we have 4c1 we have 2c1 a and another one is plus b c1 same goes for eras to power x we will have c2 plus a c2 and again it's b c2 and yes that's it equals to 0 since right hand side we can write the right hand side as like this zero times erast once again zero times erase 2x plus zero times erase to over 2x plus zero times erase to power x plus 0 times erase to power so we can compare like the component of erasover 2x with 0 and same with the component of erasover x with 0 again so we're gonna get 4 c1 plus 2 c1 a plus b c1 equals to 0 and for this case we will get c2 plus a c2 plus b c2 equals to 0 so from this we can let it call equation number one and this is equation number 2 so this will give you simply one plus a plus b equals to zero right because zero is common and similarly from here you will get four plus two a plus b equals to zero right so let call them equation number three and equation number four right so when you like subtract equation number three and equation number four you will gonna get three plus a and b and b will get cancel out so this is equal to zero so from here you will get a equals to minus of three now put this value of a here so you will get one minus three equals to minus of b so this implies b equals to two so now we can put the value of this a and b in the main equation so we will get y double prime minus three y prime plus two y equals to zero so this is our differential equation that required...

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