Help with basic differential equations?

1)

Order is the largest derivative present in the differential equation.

For yours we have:

2dy/dx + y = x - 1

Since we only have a first derivative, this is of order 1.

Now let's confirm the solution given is in fact a solution to the differential equation:

y = Ce^(-x/2) + x - 3

Differentiate to get:

dy/dx = -(C/2)e^(-x/2) + 1

Then:

2(dy/dx) + y = x - 1

==> 2[-(C/2)e^(-x/2) + 1] + [Ce^(-x/2) + x - 3] = x - 1

==> -Ce^(-x/2) + 2 + Ce^(-x/2) + x - 3 = x - 1

==> x - 1 = x - 1, so this works out

2)

x² + xy² = C

Differentiate implicitly to get:

2x + y² + 2xy(dy/dx) = 0

3)

(x² + 1)(dy/dx) + xy = 0

In order to using the integrating factor technique, the differential equation must be in the form:

(dy/dx) + f(x)y = g(x)

In your case, we are not quite in this form, so let's divide everything by x² + 1. (Note we can do this since x² + 1 will not ever be equal to 0 within the real number system):

(dy/dx) + [x / (x² + 1)]y = 0

So now we have f(x) = [x / (x² + 1)] and g(x) = 0. We now have the correct form. The integrating factor is given by:

e^( ∫ f(x) dx)

= e^( ∫ x / (x² + 1) dx)

= e^((1/2)ln(x² + 1))

= e^(ln(√(x² + 1)))

= √(x² + 1)

Multiply everything in the differential equation by the integrating factor:

(dy/dx)√(x² + 1) + [x / (x² + 1)]y = 0

Use the reverse product rule on the left side to write it as:

(√(x² + 1)y)' = 0

Integrate both sides to get:

√(x² + 1)y = 0

==> y = 0

Hence we only have y = 0 as a solution.

Rewrite this differential equation in customary kind: dy / dx = 4x - 2y dy / dx + 2y = 4x discover the values of the constants by utilising evaluating coefficients: y = ax + b dy / dx = a dy / dx + 2y = 4x a + 2(ax + b) = 4x a + 2ax + 2b = 4x 2ax + (a + 2b) = 4x 2a = 4 a = 2 a + 2b = 0 2b = -a b = -a / 2 b = -a million y = 2x - a million