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$$(t+2)x''+(t+1)x'-x=0, \qquad x(0)=0, \qquad x'(0)=2$$ and this $$ tx''-tx+4x=2e^t :: x(0)=sinh 1 ; $$

with using Derived from Laplace we get : $$ \begin{split} L(ty'') &= −s2Y'(s)−2sY(s)+y(0) \\ L(ty') &= -sY'(s)−Y(s)\\ L(ty) &= −Y'(s)\\ L(y') &= sY(s)−y(0) \end{split} $$

and when we put them in equation we get :

$$L[x](2s^2-s-2)-L'[x](s+s^2)=4s+4$$

how to solve the rest ?

how to solve this equation with laplace transform;

$$ tx''-tx+4x=2e^t :: x(0)=sinh 1 ; $$

yghboy
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    Please use MathJax for math typesetting. You can encase your math in $ a la $mx+b$ to get $mx+b$ or with $$ a la $$mx+b$$ to get $$mx+b.$$ Also avoid using " and ′ in math text. It does not render super well. For derivatives, simply use ' --- a single quote (not a smart quote either!). – Cameron Williams Jul 12 '21 at 15:07

1 Answers1

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Here is another way to approach for the sake of curiosity.

According to the substitution $u = x' + x$, we reduce the proposed ODE as follows:

\begin{align*} (t+2)x'' + (t+1)x' - x = 0 & \Longleftrightarrow (t+2)(x' + x)' - (x' + x) = 0\\\\ & \Longleftrightarrow (t+2)u' - u = 0 \end{align*}

Can you take it from here?

user0102
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  • this way is smart but is there any direct way to solve this from what i got ?@user0102 – yghboy Jul 12 '21 at 15:13
  • As far as I know, the Laplace transform is used to solve linear differential equations with constant coefficients, which is not the case. But maybe I am wrong. – user0102 Jul 12 '21 at 15:14
  • yeah but this question was in my exam and i think nobody solved it so i had to seek for answer , i want to make sure that is there any way to solve this directly or not . thanks for your help. @user0102 – yghboy Jul 12 '21 at 15:16