Question

.Find the differential equation corresponding y=ae^x+be^2x+ce^3x where a,b,c are arbitrary constant....
Please it's too urgent..... explain step by step........if u give correct answer then mark uu as brainliest...........

Answer 1

Answer:

The required differential equation is

$\bold{y'''-6y''+11y'-6y=0}$

Step-by-step explanation:

Differential equations are formed by eliminating arbitrary constants occur in the given equation.

$y=ae^x+be^{2x}+ce^{3x}$

 

$\implies\:y=e^x(a+be^x+ce^{2x})$

$\implies\:ye^{-x}=a+be^x+ce^{2x}$

Differentiate with respect to x

$ye^{-x}(-1)+e^{-x}y'=be^x+2ce^{2x}$

$\implies\:e^{-x}(-y+y')=be^x+2ce^{2x}$

$\implies\:e^{-x}(-y+y')=e^x(b+2ce^x)$

$\implies\:e^{-2x}(-y+y')=b+2ce^x$

Differentiate with respect to x

$e^{-2x}(-y'+y'')+(-y+y')e^{-2x}(-2)=2ce^x$

$\implies\:e^{-2x}(-y'+y''+2y-2y')=2ce^x$

$\implies\:e^{-2x}(y''-3y'+2y)=2ce^x$

$\implies\:e^{-3x}(y''-3y'+2y)=2c$

Differentiate with respect to x

$e^{-3x}(y'''-3y''+2y')+e^{-3x}(-3)(y''-3y'+2y)=0$

$\implies\:e^{-3x}(y'''-3y''+2y'-3y''+9y'-6y)=0$

$\implies\:e^{-3x}(y'''-6y''+11y'-6y)=0$

$\implies\:\bold{y'''-6y''+11y'-6y=0}$ which is the required differential equation.