With reduction of order, we assume a solution of the form
, with
. Then
and substituting into the ODE gives
Let
, so that
. This gives the linear ODE
This equation is also separable, so you can write
Integrating both sides with respect to
gives
Next, solve
for
by integrating both sides again with respect to
.
And finally, solve for
.
and note that
is already taken into account as part of
, so this is the general solution to the ODE.
Answer:
2
Step-by-step explanation:
Are u dumb or somthing????????
Consider, in ΔRPQ,
RP = R (Radius of larger circle)
PQ = r (radius of smaller circle)
We have to find, RQ, by Pythagoras theorem,
RP² = PQ²+RQ²
R² = r²+RQ²
RQ² = R²-r²
RQ = √(R²-r²
Now, as RQ & QS both are tangents of the smaller circle, their lengths must be equal. so, RS = 2 × RQ
RS = 2√(R²-r²)
Answer:
D. y = –2x – 7
Step-by-step explanation:
First, find the <em>rate</em><em> </em><em>of change</em><em> </em>[<em>slope</em>]:
-y₁ + y₂\-x₁ + x₂ = m
In this case, the y-intercept of [0, -7] is ALREADY given to you, so we would not have to go further in this case.
I am joyous to assist you anytime.