Solve the initial value problem: dy/dx = x√9 + x^2;y(-4)= 0?

Answer 1

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Answer 2

Since we have

dy/dx = x * sqrt(9 + x^2), all we have to do is take the integral of both sides. That is, we want to solve

Integral ( x * sqrt(9 + x^2))dx

I'm going to rearrange this to make the substitution obvious.

y = Integral ( sqrt(9 + x^2) x dx )

We solve this by substitution.

Let u = 9 + x^2. Then
du = 2x dx, and dividing both sides by 2, we have

(1/2) du = x dx

Note that our integral *has* x dx, and we replace it ENTIRELY by (1/2) du. We also do the other substitution, giving us:

y = Integral ( sqrt(u) (1/2) du )

Let's pull the constant out of the integral; constants tend to get in our way, and the wonderful thing about derivatives, integrals, and limits, is that we can ALWAYS pull out constants prior to taking the derivative, integral, or limilt.

y = (1/2) Integral ( sqrt(u) du)

Note that sqrt(u) is the same as u^(1/2)

y = (1/2) Integral ( u^(1/2) du )

Now, we use the reverse power rule, and integrate directly.

y = (1/2) [ (2/3) u^(3/2) ] + C

Let's simplify this a bit, to

y = (1/3) u^(3/2) + C

But, u = 9 + x^2, so

y = (1/3) [9 + x^2]^(3/2) + C

We're given that y(-4) = 0. However,

y(-4) = (1/3) [9 + (-4)^2]^(3/2) + C
y(-4) = (1/3) [9 + 16]^(3/2) + C
y(-4) = (1/3) [25]^(3/2) + C
y(-4) = (1/3) (125) + C
y(-4) = 125/3 + C

Equating that to 0 (since y(-4) = 0),

125/3 + C = 0, and C = -125/3

Therefore, your function is

y = (1/3) [9 + x^2]^(3/2) - 125/3

Let's verify by taking the derivative.

dy/dx = (1/3) (3/2) (9 + x^2)^(1/2) {2x}

(I used squiggly brackets to show that I used the chain rule)

dy/dx = (1/2) (9 + x^2)^(1/2) {2x}

Note the 2 in the 2x will cancel with the 2 in the one half; more specifically, let's merge them to see the cancellation with our own eyes.

dy/dx = (2x/2) (9 + x^2)^(1/2)
dy/dx = x (9 + x^2)^(1/2)
dy/dx = x sqrt(9 + x^2)

Now, let's verify that y(-4) = 0.

y(-4) = (1/3) [9 + (-4)^2]^(3/2) - 125/3
y(-4) = (1/3) [25]^(3/2) - 125/3
y(-4) = (1/3) [125] - 125/3
y(-4) = 125/3 - 125/3 = 0.

So this is indeed correct.

Answer 3

dy/dx = x√9 + x^2 = x(3) + x^2 = 3x + x^2
dy = (3x + x^2)dx
Integrating both sides
y = 3 (x2) / 2 + x^3 / 3 + c, c = constant of integration
y = 3/2 x^2 + 1/3 x^3 + c

y(-4) = 3/2 (-4)^2+ 1/3 (-4)^3 + c = 24 - 64/3 + c = - 8/3 + c

Geven
y(-4) = 0
- 8/3 + c = 0
c = 8/3

y = 3/2 x^2 + 1/3 x^3 + 8/3

Answer 4

you want to aspect x. to attempt this, discover 2 numbers that both upload as a lot as 5 and multiply to 4. that must be uncomplicated, it is basically a million and four! so that you've factored it to (x+4)(x+a million)=0. Now all of us understand that 2 issues higher mutually to make 0 can purely take position at the same time as both the first element is 0 or the 2d element is 0, so we've 2 recommendations: x+4=0 x=-4 or x+a million=0 x=-a million

Answer 5

dy/dx = x√9 + x^2;y(-4)= 0
y=(√9 / 2)x^2 + (1/3)x^3 + C
x=-4;
(√9 / 2)16 + (1/3)(-64) + C = 0
C = -(√9 / 2)16 + (1/3)(64)
y= (√9 / 2)x^2 + (1/3)x^3 + [-(√9 / 2)16 + (1/3)(64)]
No calculator on hand. That's just the setup.

Answer 6

Integrate both sides,
y+4 = (√9 / 2)x^2 + (1/3)x^3

y = (√9 / 2)x^2 + (1/3)x^3 - 4

Answer 7

dy/dx = x√9 + x^2;y(-4)= 0
Let u = 9+x^2
then du = 2x dx
dy/du = u^.5du
y =2/3 u^ 3/2 +C
y = 2/3 (9+x^2)^3/2 +C
0 = 2/3 (9+(-4)^2)^3/2 +C
C= -2/3(25)^3/2 =-2/3 (5^2)^3/2= -2/3*5^3 = -250/3
y = 2/3(9+x^2)^3/2 - 250/3

Answer 8

x = -3
y = 0