Answer to Question #200111 in Differential Equations for Raj Kumar

Question

A solution of Initial Value Problem

(1-t²)(d²y/dt²) + 2t(dy/dt) - 2y = 0, y(0) = 3, y'(0) = -4

is y1=t. Use the method of reduction of order to find a general solution of Initial Value Problem on the interval -1<t<1.

Answer 1

To solve

$(1-t^2)(\frac{d^2y}{dt^2}) + 2t(\frac{dy}{dt}) - 2y = 0$

such that

$y(0)=3, \text{ and } y'(0)=-4$

$\text{Let } y(t)=t v(t), \text{which gives }\frac{d y(t)}{d t}=t \frac{d v(t)}{d t}+v(t) \text{and }\frac{d^{2} y(t)}{d t^{2}}=t \frac{d^{2} v(t)}{d t^{2}}+2 \frac{d v(t)}{d t}\\ 2 t\left(t \frac{d v(t)}{d t}+v(t)\right)+\left(-t^{2}+1\right)\left(t \frac{d^{2} v(t)}{d t^{2}}+2 \frac{d v(t)}{d t}\right)-2 t v(t)=0\\ \text{Simplify:}\\ -t \frac{d^{2} v(t)}{d t^{2}}\left(t^{2}-1\right)+2 \frac{d v(t)}{d t}=0\\ \text{Let }\frac{d V(t)}{d t}=u(t), \text{which gives }\frac{d^{2} v(t)}{d t^{2}}=\frac{d u(t)}{d t} :\\ \text{Solve for }\frac{d u(t)}{d t}:\\ \frac{d u(t)}{d t}=\frac{2 u(t)}{t\left(t^{2}-1\right)}\\ \text{Simplify:}\\ \frac{d u(t)}{d t}=-\frac{2 u(t)}{-t^{3}+t}\\$

$\text{Divide both sides by }u(t):\\ \frac{\frac{d u(t)}{d t}}{u(t)}=-\frac{2}{-t^{3}+t}\\ \text{Integrate both sides with respect to }t :\\ \int \frac{\frac{d u(t)}{d t}}{u(t)} d t=\int-\frac{2}{-t^{3}+t} d t\\ \text{Evaluate the integrals:}\\ \log (u(t))=-2\left(\log (t)-\frac{1}{2} \log \left(-t^{2}+1\right)\right)+c_{1}, \text{where }c_{1} \text{is an arbitrary constant.}\\ \text{Solve for }u(t) :\\ u(t)=-\frac{e^{c_{1}}\left(t^{2}-1\right)}{t^{2}}\\ \text{Simplify the arbitrary constants:}\\ u(t)=\frac{c_{1}\left(t^{2}-1\right)}{t^{2}}\\$

$\text{Substitute back for }\frac{d v(t)}{d t}=u(t) :\\ \frac{d v(t)}{d t}=\frac{c_{1}\left(t^{2}-1\right)}{t^{2}}\\ \text{Integrate both sides with respect to t :}\\ v(t)=\int \frac{c_{1}\left(t^{2}-1\right)}{t^{2}} d t=c_{1}\left(\frac{1}{t}+t\right)+c_{2}, \text{ where }c_{2} \text{ is an arbitrary constant.}\\ \text{Substitute back for }y(t)=t v(t), \text{ which gives }v(t)=\frac{y(t)}{t}:\frac{y(t)}{t}=c_{1}\left(\frac{1}{t}+t\right)+c_{2}\\ \text{Solve for }y(t):\\ y(t)=c_{1}+c_{1} t^{2}+c_{2} t\\ \text{Solve for the unknown constants using the initial conditions:}\\ \text{Compute }\frac{d y(t)}{d t}:\\ \frac{d y(t)}{d t}=\frac{d}{d t}\left(c_{1}+c_{1} t^{2}+c_{2} t\right)\\ =2 c_{1} t+c_{2}\\$

$\text{Substitute }y(0)=3 \text{ into }y(t)=c_{1}+t^{2} c_{1}+t c_{2}:\\ c_{1}=3\\ \text{Substitute }y^{\prime}(0)=-4 \text{ into }\frac{d y(t)}{d t}=2 t c_{1}+c_{2} :\\ c_{2}=-4\\ \text{Solve the system:}\\ c_{1}=3\\ c_{2}=-4\\ \text{Substitute }c_{1}=3 \text{ and }c_{2}=-4 \text{ into }y(t)=c_{1}+t^{2} c_{1}+t c_{2}:\\ \text{The required solution is:}\\ y(t)=3 t^{2}-4 t+3\\$