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:

\(c_1=2,\ c_2=1\)

\(y(x)=2x^{\frac{3}{2}}+x^{-\frac{1}{2}}\)

\(4x^2-3y=0, y(1)=3, y'(1)=2.5,\) 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:

\(c_1=2,\ c_2=1\)

\(y(x)=2x^{\frac{3}{2}}+x^{-\frac{1}{2}}\)