# How to find the IVP of this equation?

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

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?

• 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. – Gerry Myerson Aug 2 '19 at 0:15
• @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? – Theo Bendit Aug 2 '19 at 0:49
• Yes, what you said is correct Theo. I'm sorry for the confusion because I don't understand this topic so well. – user616370 Aug 2 '19 at 0:53
• If $A$ and $B$ are undetermined constants, Mary, then so is their sum, so it can't make sense to work with $2t(A+B)$. – 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 $$c_{1,2}$$.