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If the DE $t^{2} y^{\prime \prime}-2 y^{\prime}+(3+t) y=0$ has $y 1, y 2$ as a
fundamental set of solutions and if $w(y 1, y 2)(2)=3$ , then find the
value of $w(y 1, y 2)(6)$

$t^{2} y^{\prime \prime}-2 y^{\prime}+(3+t) y=0$

$\div t^{2} \rightarrow y^{\prime \prime}-\frac{2}{t^{2}} y^{\prime}+\frac{(3+t) y}{t^{2}}=0$

$p(t)=\frac{-2}{t^{2}} \quad q(t)=\frac{(3+t)}{t^{2}}$

are cont every where except at $t =0$

by Abels Theorem $\left(w\left(y_{1}, y_{2}\right)\right)=c e^{-\int p(t) d(t)}$

$w\left(y_{1}, y_{2}\right)=c e^{-\int \frac{-2}{t^{2}} d t}=c e^{-2/t}$

$w\left(y_{1}, y_{2}\right)(2)=3 \rightarrow ce^{-1}=3 \Rightarrow c=3 e$

$w\left(y_{1}, y_{2}\right)(6)=3 e \cdot e^{-2 / t}=3 e \cdot e^{-1 / 3}=3 e^{2 / 3}$

If the functions $y 1$ and $y 2$ are linearly independent, determine under
what conditions the functions $f=a_{1} y_{1}+a_{2} y_{2}$ and $g=b 1 y 1+b 2 y 2$
also form a linearly independent set of functions where $a_{1}, a_{2}, b_{1}, b_{2}$ are
constants.

linearly Independent $\rightarrow w \neq 0$

$w\left(y_{1}, y_{2}\right)=\left|\begin{array}{ll}{y_{1}} & {y_{2}} \\ {y^\prime_{1}} & {y^\prime_{2}}\end{array}\right| \quad 2-1=1$

$=y_{1} y_{2}^{\prime}-y_{2} y_{1}^{\prime} \neq 0$

$w(f, g)=\left|\begin{array}{ll} {a_{1} y_{1}+a_{2} y_{2}} & {b_{1} y_{1}+b_{2} y_{2}}\\ {a_{1} y^\prime_{1}+a_{2} y^\prime_{2}} & {b_{1} y^\prime_{1}+b_{2} y^\prime_{2}}\end{array}\right| \quad 2-1=1$

$=\left(a_{1} y_{1}+a_{2} y_{2}\right)\left(b_{1} y^\prime_{1}+b_{2} y_{2}^{\prime}\right)-\left[\left(b_{1} y_{1}+b_{2} y_{2}\right)\left(a_{1} y_{1}^{\prime}+a_{2} y_{2}^{\prime}\right)\right]$

$a_{1}b_{1} y_{1}y^\prime_{1}+a_{1} b_{2} y_{1} y_{2}^{\prime}+ a_{2} b_{1} y_{2} y^\prime_{1}+ a_{2} b_{2} y_{2} y_{2}^{\prime}- a_{1}b_{1} y_{1} y_{1}^{\prime}- a_{2}b_{1} y_{1} y_{2}^{\prime}- a_{1} b_{2} y_{2} y_{1}^{\prime}- a_{2}b_{2} y_{2} y_{2}^{\prime}$

$=\left(a_{1} b_{2}-a_{2} b_{2}\right)\left(y_{1}y_{2}^{\prime}-y_2 y_{1}^\prime\right)$

$=y_{1} y_{2}^{\prime}-y_{2} y_{1}^\prime \neq 0$

The conditionis $a_{1} b_{1}-a_{2} b_{1} \neq 0$

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