Question

deee

Answer 1

could you be more specific?

Answer 2

you mean?

Answer 3

you want to solve
$$\Large y ' ' = 6x + 3$$

Answer 4

like is that a second order ode with nonconstant coefficent oh yes it is.. so in that case we need to find the Y_homogeneous and Y_p which can be done easier when you split up the problem

Answer 5

$$
\Large \text {let} ~u = y '\\
\Large \text {solve} \\
\Large u ' = 6x + 3
$$

Answer 6

^ wow what the heck mate

Answer 7

nuh uh we just set the problem to 0 to find the Y_homogeneous part of the problem
y''=0
r^2 = 0
r = 0
c1e^(0x)+c2e^(0x)x
c1+c2x so that part is right.. sorry sleepy mode. x.x

Answer 8

so the ending part of the answer is correct... now to find Yp... so that could be a polynomial type

Answer 9

but again I find it easier to split the problems as y''=6x and y''=3 both are polynomial types in the form of Ax+B but if I remember correctly if we already have a variable, we need to multiply it by x otherwise when we take the second derivative we won't be able to even grab a value.

Answer 10

\[\frac{d}{dx}(\frac{dy}{dx}) = 6x+3\]integrating once with respect to x\[\frac{dy}{dx} = \int\limits (6x + 3) dx\]\[dy/dx= 3x^2 + 3x + C_1\]integrating again\[\int\limits dy = \int\limits (3x^2 +3x+C_1 )dx \rightarrow y = x^3 + \frac{3x^2}{2}+C_1 x+C_2\]

Answer 11

$$ y ' ' = 6x + 3\\
\text {Let } u = y ' \\
\therefore\\
u' = 6x + 3 \\

\ u =6\frac{x^2}{2} +3x + c_1\\
\therefore \\
y' = 6\frac{x^2}{2} +3x + c_1\\
y = 3\frac{x^3}{3} +3\frac{x^2}{2} + c_1x + c_2
$$

Answer 12

-_- why not just call it integration... take antiderivatives and be done with it? The way I learned it was way different from this... my prof was like find yh first and then the yp followed by adding the two together yh+yp would be the final answer.

Answer 13

Thank you so so much guys. I just need to understand your diff techniques. =)

Answer 14

Ah no. Its just the same process. Thank you so much

Answer 15

=/ k