Question

how did they find the general solution? whats an auxilliary?

Answer 1

@sillybilly123 whenever you get a chance

sorry I can't be more helpful ;_;

Answer 2

@Nnesha

Answer 3

this is the given question right ?\[\rm y"=18\cos(3x) , ~~~y(0)=1 ,~~~y'(0)=4\]

Answer 4

yes

Answer 5

general solution we would have to do some integration?

Answer 6

\[\large\rm \frac{ d^2y }{ dx }= 18\cos(3x)\]
integrate
\[\frac{ dy }{ dx } =\frac{ 18 }{ 3 }\sin(3x)+c_1\]
\[\rm given:~~~y'(\color{Red}{0})=\color{blue}{4}\]\[\color{blue}{4}=\frac{ 18 }{ 3 }\sin(3\color{Red}{(0)})+c_1\]
solve for c_1

Answer 7

ohHHHH nice thanks!

Answer 8

dy/dx is the same as y'?

Answer 9

yes

Answer 10

sin(0) =0 so
\[c_1=4\]
\[\rm y'=6\sin(3x)+\color{green}{c_1}\]\[\rm y'=6\sin(3x)+\color{green}{4}\]
now we integrate again

Answer 11

y = -2cos 3x + 4 c

Answer 12

x is our variable and c is constant so when you integrate the constant you get
4x+c_2\[\large\rm y=-2\cos(3x)+4x+c_2\]

Answer 13

to double check the answer you can take derivative of 4x
the derivative of 4x is 4

Answer 14

oh thats right

Answer 15

\[\large\rm y=-2\cos(3x)+4x+c_2\]
use the given info y(0)=1 and solve for c2

Answer 16

\[1=-2\cos(3 \cdot 0)+4(0)+c_2\]\[1=-2(1)+c_2\]\[c_2=3\]\[::-> ~~~~\color{blue}{y=-2\cos(3x)+4x+3}\]