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1 answer:
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Answer:
y = x + 4
Step-by-step explanation:
the equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
y = - 3x + 6 ← is in slope- intercept form
with slope m = - 3
given a line with slope m then the slope of a line perpendicular to it is
= -
= -
=
, then
y = x + c ← is the partial equation of the perpendicular line
to find c substitute (3, 5 ) into the partial equation
5 = 1 + c ⇒ c = 5 - 1 = 4
y = x + 4 ← equation of perpendicular line
Answer:
y = -2*x^3 - x + 2
Step-by-step explanation:
We want to solve the differential equation:
y'' + 12*x = 0
such that:
y(0) = 2
y'(0) = -1
We can rewrite our equation to:
y'' = -12x
if we integrate at both sides, we get:
Solving that integral we can find the value of y', so we will get:
y' = -12* (1/2)*x^2 + C = -6*x^2 + C
where C is the constant of integration.
Evaluating y' in x = 0 we get:
y'(0) = -6*0^2 + C = C
and for the initial value problem, we know that:
y'(0) = -1
then:
y'(0) = -1 = C
C = -1
So we have the equation:
y' = -6*x^2 - 1
Now we can integrate again, to get:
y = -6*(1/3)*x^3 - 1*x + K
y = -2*x^3 - x + K
Where K is the constant of integration.
Evaluating or function in x = 0 we get:
y(0) = -2*0^3 - 0 + K
y(0) = K
And by the initial value, we know that: y(0) = 2
Then:
y(0) = 2 = K
K = 2
The function is:
y = -2*x^3 - x + 2