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})

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 \(y_2=x(\frac{3}{2})\)

Answers (1)

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:
\(c_1=2,\ c_2=1\)
\(y(x)=2x^{\frac{3}{2}}+x^{-\frac{1}{2}}\)
0

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