Skill Building
In Problems 1–12, use power series to solve each differential equation.
\(y^{\prime} +3xy=0\)
\(y^{\prime} -x+3xy=0\)
\(y^{\prime \prime}+y=0\)
\(y^{\prime \prime} +xy=0\)
\(y^{\prime \prime} +x^{2}y=0\)
\(y^{\prime \prime} -2xy=0\)
\(y^{\prime \prime} +x^{2}y^{\prime} +xy=0\)
\(y^{\prime \prime} +3xy^{\prime} +3y=0\)
\(y^{\prime \prime \prime} +y=0\)
\(y^{\prime \prime \prime} -xy=0\)
\((1+x^{2})y^{\prime \prime} -4xy^{\prime} +6y=0\)
\((x^{2}+2)y^{\prime \prime} -3xy^{\prime} +4y=0\)
In Problems 13–22:
\(y^{\prime \prime} +xy^{\prime} +y=0; \quad y(0)=1, \quad y^{\prime}(0)=0\)
\(y^{\prime \prime} -2xy^{\prime} +y=0; \qquad y(0)=2, \quad y^{\prime} (0)=1\)
\(y^{\prime \prime} -(\sin x)y=0; \qquad y(0)=0, \quad y^{\prime} (0)=1\)
\(y^{\prime \prime} +(\cos x)y=0; \qquad y(0)=0, \quad y^{\prime} ( 0) =1\)
\(y^{\prime \prime} +y^{\prime} +e^{x}y=0; \qquad y( 0) =2, \quad y^{\prime} ( 0) =1\)
1094
\(y^{\prime \prime} +(3+x)y=0; \qquad y(0)=1, \quad y^{\prime} (0)=0\)
\(y^{\prime \prime} +x^{2}y=0; \qquad y(0)=0, \quad y^{\prime} (0)=2\)
\(y^{\prime \prime} -3x^{2}y^{\prime} +2xy=0; \qquad y(0)=1, \quad y^{\prime }(0)=1\),
\(y^{(4)}-\ln (1+x)y=0; \qquad y(0)=1\), \(y^{\prime} (0)=1, \quad y^{\prime \prime} (0)=0, \quad y^{\prime \prime \prime} (0)=0\)
\(y^{\prime \prime \prime} +4y^{\prime \prime} +2y^{\prime} -x^{3}y=0; \qquad y(0)=1, \quad y^{\prime} (0)=1, \quad y^{\prime \prime} ( 0) =0\)
Applications and Extensions
Exact and Series Solutions
Exact and Series Solutions
Challenge Problems
Age of the Earth's Crust. Uranium has a half-life of \(4.5\times 10^{9}\) years. The decomposition sequence is very complicated, producing a very large number of intermediate radioactive products, but the final product is an isotope of lead with an atomic weight of 206, called uranium lead.
Assuming that the change from uranium to lead is direct, show that \(u=u_{0}e^{-kt}\), \(\ l=u_{0}(1-e^{-kt})\), where \(u\) and \(l\) denote the number of uranium and uranium lead atoms, respectively, present at time \(t\). That is, assume \(\dfrac{du}{dt} = -ku\), where \(k > 0\) is a constant, and \(l = u_0 - u\).
We can measure the ratio \(r=\dfrac{l}{u}\) in a rock, and if it is assumed that all the uranium lead came from decomposition of the uranium originally present in the rock, we can obtain a lower bound for the age of Earth's crust. Currently, \(r \approx 0.054\).