show all steps and calculus justification formative homework help!!

Question

show all steps and calculus justification formative homework help!!

Consider the differential equation dy/dx =x^2y^2with a particular solution y=f(x) having an initial condition y(−3) = 1 . Use the equation of the line tangent to the graph of f at the point (−3, 1) in order to approximate the value of f(−2.8) .

 

Given the differential equation dy/dx=-x/y^2 find the particular solution, y=f(x) , with the initial condition f(−6) = −3 .

 

What is the particular solution to the differential equation dy/dx =3 cos(x)y with the initial condition y(pi/2)=-2?

Answer 1

1) is straightforward because, since we are only asked for a tangent line approximation, we don't even have to solve the differential equation (though that wouldn't be too hard either because it is easily separable). But we know the particular solution goes through (-3 , 1) and we can plug those values for x and y into the equation for dy/dx to calculate the slope of the tangent to the curve at x = -3. So dy/dx | _(-3,1) = 9 and f(-2.8) ~ 1 + 9(.2) = 2.8

2) Separate the variables: y²dy = -xdx

Integrate both sides: 1/3y³ = - 1/2x² + C

Use initial condition to solve for C: -9 = - 18 + C , C = 9. 1/3y³ = - 1/2x² + 9 f(x) = ³√(-3/2x² + 27)

3) Separate variables: 1/y dy = 3cos(x) dx

Integrate: ln |y| = 3sinx + C

Solve for y: |y| = C₁e^3sinx.

Solve for C₁: |-2| = C₁e^3sin(pi/2). C₁ = 2e^-3.

|y| = 2e^{3sinx - 3}

y = - 2e^{3sinx - 3}

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Answer 2

These are all variables separable problems. I will show you how to one; then you work the rest.

dy/dx=x²y² => dy/y² = x²dx

-1/y = (x³/3) + C

Now you plug in the initial conditions to evaluate C

Since you have dy/dx and a point, you can write the equation of the tangent line and go from there.

If you work on these yourself and have specific questions, make a comment and I will try to help you further.