Question

If $y=\lambda_{1} e^{ax}+\lambda_{2} xe^{bx}$ where $\lambda_{1} ,\lambda_{2}$ are arbitrary constants; is general solution of $\dfrac{d^{2}y}{dx^{2}} -2\dfrac{dy}{dx} +y=0$ then the value of $\dfrac{a}{b}$ is:

Answer 1

In this question it is given that we have to find the value of $\dfrac{a}{b}$, where $y=\lambda_{1} e^{ax}+\lambda_{2} xe^{bx}$ is general solution of $\dfrac{d^{2}y}{dx^{2}} -2\dfrac{dy}{dx} +y=0$ and $\lambda_{1} ,\lambda_{2}$ are arbitrary constants. So to find the solution we have differentiate the given equation two times w.r.t ‘x’ which will lead us to a second order differential equation and after that by comparing with the above differential equation we can able to find the value of $\dfrac{a}{b}$.
Important formulas that we will be using while solution,
$\dfrac{d}{dx} \left( e^{mx}\right) =me^{mx}$,.......(1)
Where m is any arbitrary constant,
And $\dfrac{d}{dx} \left( uv\right) =u\dfrac{dv}{dx} +v\dfrac{du}{dx}$......(2)
Where u and v are the function of x.

Complete step-by-step answer:
Given equation,
$y=\lambda_{1} e^{ax}+\lambda_{2} xe^{bx}$.............(3)
Differentiating both side of the above equation w.r.t ‘x’ we get,
$\dfrac{dy}{dx} =\dfrac{d}{dx} (\lambda_{1} e^{ax}+\lambda_{2} xe^{bx})$
$\Rightarrow \dfrac{dy}{dx} =\lambda_{1} \dfrac{d}{dx} (e^{ax})+\lambda_{2} \dfrac{d}{dx} (xe^{bx})$
[ since, $\lambda_{1} ,\lambda_{2}$ are arbitrary constants that is why we take them outside the derivative]
Now by using the formula (1) and (2) the above differential equation can be written as,
$\dfrac{dy}{dx} =\lambda_{1} (ae^{ax})+\lambda_{2} \left[ x\dfrac{d}{dx} (e^{bx})+e^{bx}\dfrac{d}{dx} (x)\right]$ [where considering u=x, $v=e^{bx}$]
$\Rightarrow \dfrac{dy}{dx} =\lambda_{1} ae^{ax}+\lambda_{2} \left[ x(be^{bx})+e^{bx}\cdot 1\right]$
$\Rightarrow \dfrac{dy}{dx} =\lambda_{1} ae^{ax}+\lambda_{2} b\ xe^{bx}+\lambda_{2} e^{bx}$......(4)
Now again differentiating equation (4) w.r.t ‘x’, we get,
$\dfrac{d^{2}y}{dx^{2}} =\dfrac{d}{dx} [a\lambda_{1} e^{ax}+b\lambda_{2} e^{bx}]$
$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =a\lambda_{1} \dfrac{d}{dx} (e^{ax})+b\lambda_{2} \dfrac{d}{dx} (e^{bx})$
$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =a\lambda_{1} \left( ae^{ax}\right) +b\lambda_{2} (be^{bx})$
$\dfrac{d^{2}y}{dx^{2}} =\dfrac{d}{dx} [\lambda_{1} ae^{ax}+\lambda_{2} b\ xe^{bx}+\lambda_{2} e^{bx}]$
$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =\lambda_{1} a\dfrac{d}{dx} (e^{ax})+\lambda_{2} b\ \dfrac{d}{dx} (xe^{bx})+\lambda_{2} \dfrac{d}{dx} (e^{bx})$
$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =\lambda_{1} a(ae^{ax})+\lambda_{2} b\ \left[ x\dfrac{d}{dx} (e^{bx})+e^{bx}\dfrac{d}{dx} (x)\right] +\lambda_{2} (be^{bx})$ [using (1) and (2)]
$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =\lambda_{1} a(ae^{ax})+\lambda_{2} b\ \left[ x(be^{bx})+e^{bx}\cdot 1\right] +\lambda_{2} (be^{bx})$
$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =\lambda_{1} a^{2}e^{ax}+\lambda_{2} b^{2}xe^{bx}+2\lambda_{2} be^{bx}$.........(5)
In the question we have given that the equation (3) is the solution of the differential equation $\dfrac{d^{2}y}{dx^{2}} -2\dfrac{dy}{dx} +y=0$.........(6)
Now by putting the values of $y,\ \dfrac{dy}{dx}$ and $\dfrac{d^{2}y}{dx^{2}}$ in the equation (6), we get,
$\dfrac{d^{2}y}{dx^{2}} -2\dfrac{dy}{dx} +y=0$
$\Rightarrow (\lambda_{1} a^{2}e^{ax}+\lambda_{2} b^{2}xe^{bx}+2\lambda_{2} be^{bx})-2\left( \lambda_{1} ae^{ax}+\lambda_{2} bxe^{bx}+\lambda_{2} e^{bx}\right) +\left( \lambda_{1} e^{ax}+\lambda_{2} xe^{bx}\right) =0$
$\Rightarrow \lambda_{1} \left( a^{2}-2a+1\right) e^{ax}+\lambda_{2} \left( b^{2}-2b+1\right) xe^{bx}+2\lambda_{2} \left( b-1\right) e^{bx}=0$
Now since as we know that, $x^{2}-2xy+y^{2}=\left( x-y\right)^{2}$,
So by using this identity the above equation can be written as,
$\lambda_{1} \left( a-1\right)^{2} e^{ax}+\lambda_{2} \left( b-1\right)^{2} xe^{bx}+2\lambda_{2} \left( b-1\right) e^{bx}=0$
Now the from above equation we can say that if all the terms becomes zero then the above equation satisfied, but exponential function cannot be zero and also $\lambda_{1} ,\lambda_{2}$ cannot be zero because if $\lambda_{1} ,\lambda_{2}$ becomes zero then the given function becomes zero, i.e, y=0, which is not possible.
Therefore we can write the above equation satisfies if their coefficient becomes zero.
i.e, $\left( a-1\right)^{2} =0$, $\left( b-1\right)^{2} =0$ and $\left( b-1\right) =0$
Which implies,
a=1, b=1
Therefore the value of $\dfrac{a}{b} =\dfrac{1}{1} =1$

Note: While solving this type of question you need to know that while derivative if a constant coefficient is there then by the rule we can take it outside i.e, \dfrac{d}{dx} \left\\{ af\left( x\right) \right\\} =a\dfrac{d}{dx} \left\\{ f\left( x\right) \right\\}
Also in while solving we have used the derivative of x, so for this another formula you need to know which is $\dfrac{d}{dx} \left( x^{n}\right) =nx^{n-1}$
Since here n=1, that is why derivative of x is 1.