Find constants a , b, and c such that the function y = ax2 + bx + c satisfies the differential equation y′′ + y′ − 2y = 4x2.

a = -2, b = -2, c = 1.

Step-by-step explanation:

We know that:

y = a*x^2 + b*x + c.

Then:

y' = 2*a*x + b

y'' = 2*a.

Then we can write the equation:

y′′ + y′ − 2y = 4x^2.

as:

2*a + 2*a*x + b - 2*(a*x^2 + b*x + c) = 4*x^2

To solve this, the first step is moving all the terms to the same side:

2*a + 2*a*x + b - 2*a*x^2 - 2*b*x - 2*c - 4*x^2 = 0

Now let's group terms with the same power of x.

(-4 - 2*a)*x^2 + (-2*b + 2*a)*x + (-2c + b) = 0

Now, this must be zero for all the values of x, then the parenthesis must be equal to zero:

-4 - 2*a = 0

-2*b + 2*a = 0

-2*c + b = 0

From the first equation:

-4 - 2*a = 0

-4 = 2*a

-4/2 = a

a = -2

And from the second equation:

-2*b + 2*a = 0

2*a = 2*b

a = b

b = -2.

From the third equation we have:

-2c + b = 0

-2c = b

c = b/-2 = -2/-2 = 1.

c = 1

Then our equation is:

y = -2*x^2 - 2*x + 1.

ur right ive taken the test

step-by-step explanation:

ιт ιѕ 6/23

step-by-step explanation:

нσρє ι ¢συℓ∂ нєℓρ!