Differential equation

Here's my solution:

It should be checked (it's different from your suggested result).

.

Determine the solution to the differential equation 2y '+ y = x2 passing through the point (-1; -3).

I hope you can go from here!.

y(x) = c_1 e^(-x/2)+x^2-4 x+8

or:y'(x) = x^2/2-(y(x))/2

The original answer is correct. The mistake in the answer above is at the point where the -1 and -3 are substituted to find the value of k. The 1, in brackets on the rhs, should be -1.

Here's an alternative method of solution.

The solution consists of a Complementary Function (CF) which is the general solution of the homogeneous equation 2dy/dx + y = 0, plus a Particular Integral (PI), which is any solution of the complete equation 2dy/dx + y = x^2.

The CF is easily found to be A.exp(-x/2), where A is an arbitrary constant.

There are several methods for calculating a PI, easiest is probably by trial solution.

Let   $$y=ax^2+bx+c,$$  so  $$y'=2ax+b$$ .

Substitute into the differential equation, and we have

$$2(2ax+b)+(ax^2+bx+c)\equiv x^2$$ ,

or,    $$ax^2+(4a+b)x+(2b+c)\equiv x^2.$$

Now equate coefficients across the identity.

$$a=1,$$

$$4a+b=0,$$  so,  $$b=-4,$$

$$2b+c=0,$$  so, $$c=8.$$

That means the general solution of the original ode is    $$y = A\exp(-x/2)+x^2-4x+8,$$

and substituting the -1 and -3 gets $$A = -16 \exp(-1/2)\approx -9.70449.$$