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  • Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b. y = a e^3x + b e^-2 x

3. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

       y = ae^{3x} + b e^{-2x}

Answers (1)

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Given equation is
y = ae^{3x} + b e^{-2x}                    -(i)    
Differentiate both the sides w.r.t x
\frac{d\left ( y \right )}{dx}=\frac{d(ae^{3x}+be^{-2x})}{dx}
y^{'}=\frac{dy}{dx}= 3ae^{3x}-2be^{-2x}\\ \\                  -(ii)
Now, again differentiate w.r.t. x
y^{''}= \frac{d^2y}{dx^2} = 9ae^{3x}+4be^{-2x}            -(iii)
Now, multiply equation (i) with 2 and add equation (ii)
2(ae^{3x}+be^{-2x})+(3a-2be^{-x}) = 2y+y^{'}\\ 5ae^{3x} = 2y+y^{'}\\ ae^{3x}= \frac{2y+y^{'}}{5}  -(iv)
Now,  multiply equation (i) with 3 and subtract from equation (ii)
3(ae^{3x}+be^{-2x})-(3a-2be^{-x}) = 3y-y^{'}\\ 5be^{-2x} = 3y-y^{'}\\ be^{-2x}= \frac{3y-y^{'}}{5}-(v)
Now, put values from (iv) and (v) in equation (iii)
y^{''}= 9.\frac{2y+y^{'}}{5}+4.\frac{3y-y^{'}}{5}\\ \\ y^{''}= \frac{18y+9y^{'}+12y-4y^{'}}{5}\\ \\ y^{''}= \frac{5(6y-y^{'})}{5}=6y-y^{'}\\ \\ y^{''}+y^{'}-6y=0

Therefore, the required differential equation is  y^{''}+y^{'}-6y=0

Posted by

Gautam harsolia

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