\begin{cases} \dfrac{\mathrm{d}^2y}{\mathrm{d}t^2} + 4y = 2t \\ y(0) = 1 \\ y'(0) = 2 \end{cases}

This is incorrect, but I'm unsure how to find the homogeneous solution $Y_h(t)$. I tried to do:

$Y_h(t) = 2t(A + B)$

$Y_p(t)$ means particular solution: $Y_p(t) = 2t^2(A + B)$

Then, I do the first derivative and second derivatives:

$Y'_p(t) = 4t(A + B)$

$Y''_p(t) = 4(A + B)$

Once I plug back the derivatives into the original equation, I got:

$4(A + B) + 8t^2(A + B) = 2t$

Could someone please kindly explain what I did incorrectly?

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    $\begingroup$ Incomprehensible. What's an IVP? "IVP of this equation" – of what equation? What is $Y(h)$ supposed to mean? How is $h$ related to $t$? How does $p$ come into it? Why do you write $2At+2Bt$ instead of $2(A+B)t$? How do you get $4A+4B+8At^2+8Bt^2=2t$? This is a real dog's breakfast of a question. $\endgroup$ – Gerry Myerson Aug 2 '19 at 0:15
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    $\begingroup$ @GerryMyerson I think that $Y(h)$ is supposed to be the general solution to the homogeneous equation $y'' + y = 0$ (i.e. better written as $Y_h(t)$). Similarly, $Y(p)$ is, as Mary says, a particular solution to the given IVP (should be $Y_p(t)$). Could you confirm this is case Mary? $\endgroup$ – Theo Bendit Aug 2 '19 at 0:49
  • $\begingroup$ Yes, what you said is correct Theo. I'm sorry for the confusion because I don't understand this topic so well. $\endgroup$ – user616370 Aug 2 '19 at 0:53
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    $\begingroup$ If $A$ and $B$ are undetermined constants, Mary, then so is their sum, so it can't make sense to work with $2t(A+B)$. $\endgroup$ – Gerry Myerson Aug 2 '19 at 0:53

Given that the OP is using $y(p)$ to represent the particular solution, I assume that $y(h)$ represents the solution of the homogeneous equation-- sometimes (in my experience) denoted $y_h(t)$. By definition, the homogeneous solution must satisfy $$ \frac{d^2}{dt^2}y_h(t)+4y_h(t) = 0.$$

This is a linear, second order differential equation, so we assume that $y_h(t) = \exp(\lambda t)$ to arrive at the characteristic equation $$ (\lambda^2 +4)\exp(\lambda t) = 0, $$ which can hold only if $ \lambda^2 +4 =0$. Thus, we arrive at $\lambda = \pm 2i$, and $$ y_h(t) = c_1 \cos(2t) + c_2\sin(2t)$$ with unknown coefficients $c_{1,2}$ to be determined from the initial conditions.

Now, to determine the non-homogeneous solution, $y_p(t)$ it appears as though OP is looking to use the undetermined coefficients method. The non-homogeneous equation is $$ \frac{d^2}{dt^2}y_p(t)+4y_p(t) = 2t.$$ It is not so hard to see that the right hand side of the non-homogeneous equation is a first degree polynomial. Thus, we guess that the particular solution will also be first degree polynomial and write $$ y_p(t) = At+b $$. Inserting this guess into the non-homogeneous equation, and noticing that the second derivative kills the first degree polynomial, gives $$ 4y_p(t) = 4(At+B) = 2t$$ for $A,B$ unknown constants.

Collecting common terms in the above equation yields the non-homogeneous solution. Finally, evaluation of $y(t) = y_p(t)+y_h(t)$ at the initial conditions ($t=0$) gives the values of the unknown constants $c_{1,2}$.

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