Respuesta :
Answer:
[tex]y(x)\ =\ \sqrt{x}[C_1cos\dfrac{\sqrt{3}}{2}logx+C_2sin\dfrac{\sqrt{3}}{2}logx][/tex]
Step-by-step explanation:
Given differential equation is
[tex]x^2y"-xy'+y=0[/tex] (1)
Let's assume that
[tex]x=e^t[/tex]
[tex]=>\ t\ =\ logx[/tex]
then,
[tex]\dfrac{dx}{dt}=e^t[/tex]
[tex]and\ \dfrac{d^2x}{dt^2}=e^t[/tex]
We can write,
[tex]\dfrac{dy}{dx}=\dfrac{dy}{dt}.\dfrac{dt}{dx}[/tex]
[tex]=e^{-t}\dfrac{dy}{dt}[/tex]
Similarly,
[tex]\dfrac{d^2y}{dt^2}=\dfrac{d^2y}{dt^2}.\dfrac{dt^2}{dx^2}[/tex]
[tex]=e^{-2t}.\dfrac{d^2y}{dt^2}[/tex]
Putting these values in equation (1), we will get
[tex]e^{2t}.e^{-2t}.\dfrac{d^y}{dt^2}-e^t.e^{-t}\dfrac{dy}{dt}+y=0[/tex]
[tex]=>\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+y=0[/tex]
So, the characteristics equation can be given as
[tex]D^2-D+1=0[/tex]
[tex]=>D\ =\ \dfrac{1+\sqrt{1-4}}{2}\ or\ \dfrac{1-1\sqrt{1-4}}{2}[/tex]
[tex]=>D=\ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}\ or\ \dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}[/tex]
Hence, the general solution of the equation can be give by
[tex]y(t)\ =\ e^{\dfrac{t}{2}}[C_1cos\dfrac{\sqrt{3}}{2}t+C_2sin\dfrac{\sqrt{3}}{2}t][/tex]
Now, by putting the value of t in above solution, we will have
[tex]y(x)\ =\ e^{\dfrac{1}{2}logx}[C_1cos\dfrac{\sqrt{3}}{2}logx+C_2sin\dfrac{\sqrt{3}}{2}logx][/tex]
[tex]y(x)=\ \sqrt{x}[C_1cos\dfrac{\sqrt{3}}{2}logx+C_2sin\dfrac{\sqrt{3}}{2}logx][/tex]
Hence, the solution of above given differential equation can be given by
[tex]y(x)=\ \sqrt{x}[C_1cos\dfrac{\sqrt{3}}{2}logx+C_2sin\dfrac{\sqrt{3}}{2}logx][/tex]
