Question

How do I solve 2y'' - 3y' - 2y = 13 - 2x^2? Please, its a general solution.

Answer 1

Isn't that 13 - 4 on the right side?

Answer 2

2x^2 pls

Answer 3

?

Answer 4

find general solution?

Answer 5

Yes Sam

Answer 6

try let \[y\text{ = }e^{k x}\]

Answer 7

OK Sam, am very new to this n hope you help with the answer

Answer 8

\[2 y''-3 y'-2 y=13-2^2\]
\[2 \frac{d^2\text{}}{dx^2}\left(e^{k x}\right)-3 \frac{d\text{}}{dx}\left(e^{k x}\right)-2 e^{k x}\text{ = }0\]
sub \[\text{}\frac{d^2\text{}}{dx^2}\left(e^{k x}\right)\text{ = }k ^2 e^{k x}\text{ and }\frac{d\text{}}{dx}\left(e^{k x}\right)\text{ = }k e^{k x}\]
\[2k ^2 e^{k x}-3 k e^{k x}-2 e^{k x}\text{ = }0\]
factor
\[\left(2 k^2-3 k -2\right) e^{k x}\text{ = }0\]
\[2 k ^2-3k -2\text{ = }0\]
\[k=-\frac{1}{2}\text{ or }k =2\]

Answer 9

Sam how do I edit my question?

Answer 10

there's a button beside your question

Answer 11

= 13 - 2x^2. 2 ex raised to power 2 pls

Answer 12

Thanx Sam, I ve done that. Pls I have an exam on that and even IVPs

Answer 13

= 13 - 2x^2. 2 ex raised to power 2 pls

Answer 14

Sam how do I edit my question?

Answer 15

OK Sam, am very new to this n hope you help with the answer

Answer 16

find out the complimentary function and the particular integral......
add the two.......
y=C.F+P.I

Answer 17

Okay Thanks

Answer 18

did you tried wolfram?

Answer 19

no i havent its been tough,,