IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

karthik2007

That method is very easy to apply actually. You don't have to do integration or differentiation, and you can solve equations involving fourth order Diff. coefficients easily. I can post the method too, if ppl are interested

elastiboysai

ya post it.

I sort of understood what amrish was saying.

I saw d ans n did some golmaal substitution to get the ans.

elastiboysai

K. i looked at the ans n figd dis out

consider k=d^2y/dx^2 +2dy/dx

k=dk/dx

so k=ae^x

now plug in n solve..

magiclko

y"' + y" -2y' =0

here we can have an operator D, where, Dy = y' , D²y = y' .....Dⁿ = nth derivative of n wrt x

now the equation can be written as

(D³ + D² -2 D ) y=0

=> D³ + D² -2 D =0

=> D = 0 , -2,1

thus the roots of the auxillary equation are real and distinct....

so the solution will be

y = Ae^0x+ Be^-2x+ Ce^1x

=> y = A +Be^-2x+ Ce^x

magiclko

and the concept cums frm here.....

............................................................................

A second order linear homogeneous ordinary differential equation with constant coefficients can be expressed as

a₂y" + a₁y' +a₀ y =0

This equation implies that the solution is a function whose derivatives keep the same form as the function itself and do not explicitly contain the independent variable x, since constant coefficients are not capable of correcting any irregular formats or extra variables. An elementary function which satisfies this restriction is the exponential function

Substitute the exponential function

into the above differential equation, the characteristic equation of this differential equation is obtained

a_2² + a₁ +a₀ =0

This characteristic equation has two roots

_¹ and

_²

2nd Order Linear Homogeneous ODE with Constant Coefficients:
Characteristic Equation:
Solutions of Characteristic Equation ,		General Solution
1
2
3

karthik2007

For example you have a DE of the form d⁴y/dx⁴+d³y/dy³+2d²y/dx²+dy/dy+y = 0

Now, what you do is, you replace the operator dⁿ/dxⁿ by an algebraic variable, called D

Hence, the equation reduces to:

y(D⁴+D³+2D²+D+1) = 0.

Now, you take the equation in the bracket, ie, the one in terms of D, and you equate it to zero and find its roots.

If all roots are real, and distinct, the solution is of the form:

y = Ae^px+Be^qx+C^wx+De^gx, where the coefficients of x are the roots of the equation. Here we have four terms because we have four roots.

If the roots are complex, of the form a

ib

Then, solution is of the form y = e^ax(Acosbx+Bsinbx)

Hence, if two roots are real, and two imaginary, you have to write the solution separately for the two imaginary roots, and for the two real roots, and then add them to get the final solution.

There are actually many more varieties, but this should suffice to handle any type of DE at JEE level.

Ask & Discuss Questions with Community & Experts