Question
Question: The differential equation of the family of curves \[y = a{e^x} + bx{e^x} + c{x^2}{e^x}\], where \[a,...
The differential equation of the family of curves y=aex+bxex+cx2ex, where a,b and c are arbitrary constants, is:
A. y′′′+3y′′+3y′+y=0
B. y′′′+3y′′−3y′−y=0
C. y′′′−3y′′−3y′+y=0
D. y′′′−3y′′+3y′−y=0
Solution
In order to solve this question we have to differentiate the family of curve y=aex+bxex+cx2ex and there are three arbitrary constants in the equation and we have to differentiate this equation three times and eliminate all the arbitrary constant from all the three equation. And final equation in terms and x, and y.
Complete step by step answer:
We have given a family of curves and we have to find the differential equation of this curve.
In this y=aex+bxex+cx2ex equation of curve we have given three arbitrary constants so we have to differentiate this equation three times.
dxdy is represented by y′.
⇒y=aex+bxex+cx2ex
On differentiating with respect to x,
y′=aex+b(xex+ex)+c(x2ex+2xex)
On simplifying this equation.
y′=aex+bxex+bex+cx2ex+c2xex
Now expressing this term in terms of y.
y′=y+bex+c2xex
Now, again differentiating with respect to x.
y′′=y′+bex+2c(xex+ex)
On further solving
y′′=y′+bex+2cxex+2cex
Now putting the value in terms of y′ and y in this equation.
y′′=2y′−y+2cex
Again differentiating with respect to x.
y′′′=2y′′−y′+2cex
Now putting the value of 2cex from the last obtained equation.
y′′′=2y′′−y′+y′′−2y′+y
On further solving
∴y′′′−3y′′+3y′−y=0
Hence, the differential equation of family of curve y=aex+bxex+cx2ex is y′′′−3y′′+3y′−y=0.
Therefore, option “D” is the correct answer.
Note: To solve these types of questions students must know the multiplication rule of the differentiation and students must know how they solve the equation. There are many places where students often make mistakes. To find a differential equation of a curve having n number of arbitrary constants then we have to differentiate that equation n times.