Solveeit Logo

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+cx2exy = a{e^x} + bx{e^x} + c{x^2}{e^x}, where a,ba,b and cc are arbitrary constants, is:
A. y+3y+3y+y=0{y^{\prime \prime \prime }} + 3y'' + 3y' + y = 0
B. y+3y3yy=0{y^{\prime \prime \prime }} + 3y'' - 3y' - y = 0
C. y3y3y+y=0{y^{\prime \prime \prime }} - 3y'' - 3y' + y = 0
D. y3y+3yy=0{y^{\prime \prime \prime }} - 3y'' + 3y' - y = 0

Explanation

Solution

In order to solve this question we have to differentiate the family of curve y=aex+bxex+cx2exy = a{e^x} + bx{e^x} + c{x^2}{e^x} 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+cx2exy = a{e^x} + bx{e^x} + c{x^2}{e^x} equation of curve we have given three arbitrary constants so we have to differentiate this equation three times.
dydx\dfrac{{dy}}{{dx}} is represented by yy'.
y=aex+bxex+cx2ex\Rightarrow y = a{e^x} + bx{e^x} + c{x^2}{e^x}
On differentiating with respect to x,
y=aex+b(xex+ex)+c(x2ex+2xex)y' = a{e^x} + b\left( {x{e^x} + {e^x}} \right) + c\left( {{x^2}{e^x} + 2x{e^x}} \right)

On simplifying this equation.
y=aex+bxex+bex+cx2ex+c2xexy' = a{e^x} + bx{e^x} + b{e^x} + c{x^2}{e^x} + c2x{e^x}
Now expressing this term in terms of y.
y=y+bex+c2xexy' = y + b{e^x} + c2x{e^x}
Now, again differentiating with respect to x.
y=y+bex+2c(xex+ex)y'' = y' + b{e^x} + 2c\left( {x{e^x} + {e^x}} \right)
On further solving
y=y+bex+2cxex+2cexy'' = y' + b{e^x} + 2cx{e^x} + 2c{e^x}

Now putting the value in terms of yy' and yy in this equation.
y=2yy+2cexy'' = 2y' - y + 2c{e^x}
Again differentiating with respect to x.
y=2yy+2cexy''' = 2y'' - y' + 2c{e^x}
Now putting the value of 2cex2c{e^x} from the last obtained equation.
y=2yy+y2y+yy''' = 2y'' - y' + y'' - 2y' + y
On further solving
y3y+3yy=0\therefore y''' - 3y'' + 3y' - y = 0
Hence, the differential equation of family of curve y=aex+bxex+cx2exy = a{e^x} + bx{e^x} + c{x^2}{e^x} is y3y+3yy=0y''' - 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.