Verify that the given functions form a basis of solutions of the given equation and solve the given initial value problem. 4x^2-3y=0, y(1)=3, y'(1)=2.5, the basis of solution are y_1=x^{-frac{1}{2}} and y_2=x(frac{3}{2})

Question
Verify that the given functions form a basis of solutions of the given equation and solve the given initial value problem.
$$4x^2-3y=0,\ y(1)=3,\ y'(1)=2.5,$$ the basis of solution are $$y_1=x^{-\frac{1}{2}}$$ and $$y_2=x(\frac{3}{2})$$

2021-01-05
Given:
$$4x^2-3y=0, y(1)=3, y'(1)=2.5,$$ the basis of solution are $$y_1=x^{-\frac{1}{2}}$$ and $$y_2=x(\frac{3}{2})$$
$$4x^2y''-3y=0$$
Let $$x=e^t$$
$$\Rightarrow\frac{dx}{dt}=e^t$$
$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=e^{-t}\frac{dy}{dt}$$
$$\Rightarrow e^t\frac{dy}{dx}=\frac{dy}{dt}$$
$$\Rightarrow x\frac{dy}{dx}=\frac{dy}{dt}$$
Similarly $$x^2\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}-\frac{dy}{dt}$$
$$4x^2y''-3y=0$$
$$\Rightarrow\frac{4d^2y}{dt^2}-\frac{4dy}{dt}-3y=0$$
Auxiliary equation is given by:
$$4m^2-4m-3=0$$
$$\Rightarrow m=\frac{3}{2},\frac{-1}{2}$$
$$y(t)=c_1e^{\frac{3}{2t}}+c_2e^{\frac{-1}{2t}}$$
put $$t=\ln x\ \ (x=e^t)$$
$$\Rightarrow y(x)=c_1x^{\frac{3}{2}}+c_2x^{-\frac{1}{2}}$$
So any solution is a linear combination of
$$x^{\frac{3}{2}}$$ and $$x^{\frac{1}{2}}$$
also $$x^{\frac{3}{2}}$$ and $$x^{\frac{1}{2}}$$ are linearly independent so $$[x^{\frac{3}{2}},x^{\frac{1}{2}}]$$ forms a basis for the solution of $$4x^2y''-3y=0$$
Now, $$y(x)=c_1x^{\frac{3}{2}}+c_2x^{\frac{1}{2}}$$
given $$y(1)=3,\ y'(1)=25$$
$$y(1)=3$$
$$\Rightarrow3=c_1+c_2$$
$$y'(x)=\frac{3}{2}c_1x^{\frac{1}{2}}-\frac{1}{2}c_2x^{\frac{3}{2}}$$
$$y'(x)=2.5$$
$$\Rightarrow2.5=\frac{3}{2}c_1-{\frac{1}{2}}c_2$$
$$\Rightarrow5=3c_1-c_2$$
Solving we get:
$$c_1=2,\ c_2=1$$
$$y(x)=2x^{\frac{3}{2}}+x^{-\frac{1}{2}}$$

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