Author Topic: Q4 TUT 0701  (Read 1047 times)

Mikhail Pavlovskiy

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Q4 TUT 0701
« on: October 26, 2018, 03:04:22 PM »
Question
Verify that the given functions $y_{1}$ and $y_{2}$ satisfy the corresponding homogeneous equation; then find a particular solution of the given inhomogeneous equation.
$t y'' - (1 + t)y' + y = t^2 e^{2t}$, $t>0$
$y_{1} = 1 + t$, $y_{2} = e^t$

Solution:
1) Let's check if $y_{1}$ is a solution:
$y_{1} = 1+t$, $y_{1}' = 1$, $y_{1}'' = 0$ - derivatives of $y_{1}$
Substitute them into homogenuous(left) part of our equation:
$t * 0 - (1+t) * 1 + (1+t) = 0 \Rightarrow y_{1}$ is a solution

2) Let's check if $y_{2}$ is a solution:
$y_{2} = e^t$, $y_{2}' = e^t$, $y_{2}'' = e^t$ - derivatives of $y_{2}$
Substitute them into homogenuous(left) part of our equation:
$t * e^t - (1+t) * e^t + e^t = 0 \Rightarrow y_{2}$ is a solution
   

3) Let's find particular solution of our equation:
$y_{p} = y_{1}*u + y_{2}*v$
where $u = - \int \frac{y_{2}*r(x)}{W(y_{1},y_{2})}$,
$v = \int \frac{y_{1}*r(x)}{W(y_{1},y_{2})}$. Since $r(x) = t^2 e^{2t} / t$ and $W(y_{1},y_{2})= t e^t$, we get that $u = - \frac{1}{2} e^{2t}$
and $v = t e^t$. Then the particular after substitution $y_{1}, y_{2}, u, v$ into $y_{p}$, we get:
$y_{p} =  \frac{1}{2} (t-1) e^{2t} $

Victor Ivrii

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Re: Q4 TUT 0701
« Reply #1 on: October 26, 2018, 03:36:51 PM »
Do not post before problems are posted!
Never use * as a sign of multiplication, because it is reserved for a different operation-convolution. For multiplication you should use $\cdot$ (\cdot) or $\times$ (\times) or no sign at all

Xiaoyuan Wang

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Re: Q4 TUT 0701
« Reply #2 on: October 26, 2018, 08:48:24 PM »
answer