Give the general solution to y'' - 3y' + 2y = 4x

note: y' is often written as dy/dx, this doesn't change the question at all so no need to worry. Same for y'' and d²y/dx².All second-order differential equations that don't have 0 on the right hand side have two parts to them, the complementary function (CF) and the particular integral (PI).Part 1: CFThe complementary function comes from solving the left hand side equal to 0. So we have to solve: y''-3y'+2y=0. First thing to do is to create an auxiliary equation (these are all just names, don't worry about them) by changing the derivates of y to some power of a variable like m, so y'' becomes m^2, y' becomes m and y becomes a constant. We insert this into the equation above and end up with:m^2-3m+2=0, which we can factorise to (m-1)(m-2)=0 to get m=1 and m=2.Now there are 3 different complementary functions depending on the m's we get as solutions. In general for 2 distinct solutions we have e^cx and e^dx if we had m = c and m = d as solutions. In this case we'd get y = Ae^x+ Be^2x (don't forget the constants) as our complementary function! If the right hand side was equal to 0, we'd be done here!Part 2: PIFor the second part we consider the right hand side, in this case we have 4x. There are quite a few different formulas you may need to remember about the particular integral but you'll know them with time. In our case 4x is a linear polynomial, there is an x and a constant. So for our PI we can write it as y = ax + b (this will always be a general version of whatever function of x is on the right hand side, with a few tweaks in certain situations). We want to then put this into the equation we're trying to solve, so first calculate the derivatives: y' = a, and y'' = 0.Putting these into the equation gets us:0 - 3a + 2(ax + b) = 4x, and expanding this out gives 2ax+2b-3a = 4x. We can solve this by comparing the coefficients of x, and the constants. The x coefficients give us that 2a = 4, so a = 2. Then the constants give us 2b-3a=0, subbing in a and rearranging shows b = 3. So plugging these back into our PI (y = ax + b) gives y = 2x + 3. The final answer is simply the sum of these two functions! Since we are finding a general solution (i.e. we have no initial conditions), the answer is y = Ae^x + Be^2x + 2x + 3.