Question

Consider the differential equation dy over dx equals 4 times quantity 2 times x plus 2 end quantity times sin of quantity x squared plus 2 times x plus pi over 2 end quantity period Part A: Find the equation of the line tangent to the solution curve at the point (0, 4). (5 points) Part B: Find the second derivative at (0, 3) and use it to determine the concavity of the solution curve at that point. Explain. (10 points) Part C: Find the particular solution y

Answer 1

It looks like the differential equation is

`(dy)/(dx) = 4 (2x+2) \sin\left(x^2 + 2x + \frac\pi2\right)`

A.
`(dy)/(dx)` gives the slope of the line tangent to the curve
`y=y(x)` at the point
`(x,y)`. At the point (0, 4), the tangent line has slope

`(dy)/(dx)\bigg|_(x=0,y=4) = 4\sin\left(\frac\pi2\right) = 4`

Then using the point-slope formula, the equation of the line is

`y - 4 = 4 (x-0) \implies \boxed{y=4x+4}`

B. Differentiate both sides of the ODE with respect to
`x`. Using the product, chain, and power rules,

`(d^2y)/(dx^2) = 8 \sin\left(x^2+2x+\frac\pi2\right) + 4(2x+2)^2 \cos\left(x^2+2x+\frac\pi2\right)`

You're probably supposed to evaluate the second derivative at (0, 4), not (0, 3), since we don't know whether (0, 3) is on the solution curve (yet). At this point,

`(d^2y)/(dx^2) \bigg|_(x=0,y=4) = 8 \sin\left(\frac\pi2\right) + 16 \cos\left(\frac\pi2\right) = 8`

Since 8 > 0, the solution curve is concave upward at (0, 4).

C. Using the point (0, 4) as an initial value, integrating both sides of the ODE with the fundamental theorem of calculus gives

`\displaystyle y(x) = y(0) + \int_0^x 4 (2u + 2) \sin\left(u^2 + 2u + \frac\pi2\right) \, du`

In the integral, substitute
`v=u^2+2u+\frac\pi2` and
`dv=(2u+2)\,du`.

`\displaystyle y(x) = 4 + \int_(\pi/2)^(x^2+2x+\pi/2) 4 \sin(v) \, dv`

`\displaystyle y(x) = 4 - 4 \cos(v) \bigg|_(v=\pi/2)^(v=x^2+2x+\pi/2)`

`\displaystyle y(x) = 4 - 4 \left(\cos\left(x^2 + 2x + \frac\pi2\right) - \cos\left(\frac\pi2\right)\right)`

`\displaystyle y(x) = 4 - 4 \cos\left(x^2 + 2x + \frac\pi2\right)`

which we can expand as

`\cos\left(x^2+2x+\frac\pi2\right) = \cos(x^2+2x)\cos\left(\frac\pi2\right) - \sin(x^2+2x)\sin\left(\frac\pi2\right) \\\\ ~~~~= -\sin(x^2+2x)`

Then the particular solution to the ODE is

`\boxed{y(x) = 4 - 4\sin(x^2+2x)}`

(and we see that
`x=0` only yields one value
`y=4`, so (0, 3) is indeed not on the curve)