Question

dy/dx+2y=2x^2+3

yP(x)=x^2−x+2
yP(x)= x^2+x+2
yP(x)=− x^2−x+2
yP(x)=− x^2−x−2

Answer 1

Answer on Question #79198 – Math – Differential Equations

Question

$$\mathrm{dy/dx} + 2y = 2x^2 + 3$$ $$y_P(x) = x^2 - x + 2$$ $$y_P(x) = x^2 + x + 2$$ $$y_P(x) = -x^2 - x + 2$$ $$y_P(x) = -x^2 - x - 2$$

Solution

Note that if $y_P(x)$ is a particular solution of the differential equation $\mathrm{dy/dx} + 2y = 2x^2 + 3$, then substituting $y_P(x)$ for $y$ and using the derivative of $y_P(x)$ we obtain the identity.

1) Consider $y_P(x) = x^2 - x + 2$.

$$\mathrm{dy}_D(x)/\mathrm{dx} = \frac{d(x^2 - x + 2)}{dx} = 2x - 1, \quad 2y_D(x) = \frac{2(x^2 - x + 2)}{2x^2} = 2x^2 - 2x + 4.$$

The left-hand side of our equation in this case:

$$\mathrm{dy}_D(x)/\mathrm{dx} + 2y_D(x) = 2x - 1 + 2x^2 - 2x + 4 = 2x^2 + 3.$$

We get $2x^2 + 3 = 2x^2 + 3$, hence $y_P(x) = x^2 - x + 2$ is a particular solution of the differential equation $\mathrm{dy/dx} + 2y = 2x^2 + 3$.

2) Consider $y_P(x) = x^2 + x + 2$.

$$\mathrm{dy}_D(x)/\mathrm{dx} = \frac{d(x^2 + x + 2)}{dx} = 2x + 1, \quad 2y_D(x) = \frac{2(x^2 + x + 2)}{2x^2} = 2x^2 + 2x + 4.$$

The left-hand side of our equation in this case:

$$\mathrm{dy}_D(x)/\mathrm{dx} + 2y_D(x) = 2x + 1 + 2x^2 + 2x + 4 = 2x^2 + 4x + 5.$$

We get $2x^2 + 4x + 5 \neq 2x^2 + 3$, hence $y_P(x) = x^2 + x + 2$ is not a particular solution of the differential equation $\mathrm{dy/dx} + 2y = 2x^2 + 3$.

3) Consider $y_P(x) = -x^2 - x + 2$.

$$\mathrm{dy}_D(x)/\mathrm{dx} = \frac{d(-x^2 - x + 2)}{dx} = -2x - 1, \quad 2y_D(x) = \frac{2(-x^2 - x + 2)}{2x^2} = -2x^2 - 2x + 4.$$

The left-hand side of our equation in this case:

$$\mathrm{dy}_D(x)/\mathrm{dx} + 2y_D(x) = -2x - 1 - 2x^2 - 2x + 4 = -2x^2 - 4x + 3.$$

We get $-2x^2 - 4x + 3 \neq 2x^2 + 3$, hence $y_P(x) = -x^2 - x + 2$ is not a particular solution of the differential equation $\mathrm{dy/dx} + 2y = 2x^2 + 3$.

4) Consider $y_P(x) = -x^2 - x - 2$.

$$\mathrm{dy}_D(x)/\mathrm{dx} = \frac{d(-x^2 - x - 2)}{dx} = -2x - 1, \quad 2y_D(x) = \frac{2(-x^2 - x - 2)}{2x^2} = -2x^2 - 4x - 4.$$

Left-hand side of our equation in this case:

$$\mathrm{dy}_D(x)/\mathrm{dx} + 2y_D(x) = -2x - 1 - 2x^2 - 2x - 4 = -2x^2 - 4x - 5.$$

We get $-2x^2 - 4x - 5 \neq 2x^2 + 3$, hence $y_P(x) = x^2 - x - 2$ is not a particular solution of the differential equation $\mathrm{dy/dx} + 2y = 2x^2 + 3$.

Answer: the first option is correct, $y_{P}(x) = x^{2} - x + 2$.

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