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

Hint: 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.