Question

Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. t2y'' − 2y = 5t2 − 1, t > 0, y1(t) = t2, y2(t) = t−1

Answer 1

Answer:

$y_1, y_2$ satisfies given homogenous solution and the particular solution is $\frac{5}{t^4}+\frac{1}{2t^2}-\frac{5}{2}$.

Step-by-step explanation:

Given homogeneous equation is,

$t^2y''-2y=5t^2-1$ where t>0 subject to $y_1(t)=t^2, y_2(t)=\frac{1}{t}\hfill (1)$

To verify,

• wheather $y_1, y_2$ satisfies given homogenous equation, then both will satisfy $t^2y''-2y=0$  that is,

$t^2y_{1}^{''}-2y_1=t^2\times 2-2\times t^2=0$

$t^2y_{2}^{''}-2y_2=t^2\times 2t^{-3}-2\times t^{-1}=2t^{-1}-2t^{-1}=0$

Thus $y_1$ and $y_2$ satisfies (1).

Now the wronskean,

$W(y_1, y_2)(x)=y_1y_{2}^{'}-y_{2}y_{1}^{'}=-t^2t^{-2}-2tt^{-1}=-3\neq 0$

Thus $y_1$ and $y_2$ are solution of (1) .

• Particular solution,

$Y(t)=\frac{1}{t^2D^2-2}(5t^2-1)$      where $D\equiv \frac{\partial }{\partial t}$

$=-\frac{1}{2t^2}\frac{1}{1-\frac{D^2}{t^2}}()5t^2-1$

$=-\frac{1}{2t^2}(1-\frac{D^2}{t^2})^{-1}(5t^2-1)$

$=-\frac{1}{2t^2}(1-\frac{D^2}{t^2}+........)(5t^2-1)$

$=-\frac{1}{2t^2}(5t^2-1-\frac{10}{t^2})$     sincce $D^2(5t^2-1)=10$

$=\frac{5}{t^4}+\frac{1}{2t^2}-\frac{5}{2}$

Hence the result.