How can we solve the differential equation y y 14 y 4 e^{x^2} using the method of undetermined coefficients?

To solve the differential equation y” + 14y’ + 4y = e^{x^2} using the method of undetermined coefficients, we should follow these steps:

1. Find the Complementary Solution (y_c)

The complementary solution is found by solving the associated homogeneous equation:

y” + 14y’ + 4y = 0

We can start by finding the characteristic equation:

r² + 14r + 4 = 0

Using the quadratic formula, we find:

r = rac{-b
ightarrow 14 frac{14}{2}
ightarrow 13.0r = rac{-14
ightarrow rac{rac{14}{2}
ightarrow 6.5} + rac{rac{4}{2}
ightarrow 2
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Thus, our complementary solution is a combination of exponentials:

y_c = C₁e^{r₁x} + C₂e^{r₂x}

2. Find the Particular Solution (y_p)

Next, we look for a particular solution. Since our non-homogeneous term is e^{x^2}, we will assume a particular solution of the form:

y_p = Ae^{x^2} + Bxe^{x^2}

where A and B are the undetermined coefficients we must solve for.

3. Substitute and Solve

We now need to compute y_p‘ and y_p”:

• The first derivative:

y_p‘ = (2Ax + B)e^{x^2} + (Ax + Bxe^{x^2}) · 2x

• The second derivative:

y_p” = (2A + 2Bx)e^{x^2} + [(2Ax + B)e^{x^2} + 2x(Axe^{x^2})] · 2

Substituting y_p, y_p‘, and y_p” back into the original equation allows us to collect like terms.

4. Equate coefficients

After all substitutions and simplifications, you will obtain a system of equations with A and B. Solve these equations to find the values of A and B.

5. General Solution

Once you have both the complementary and particular solutions, you can express the general solution as:

y = y_c + y_p

In conclusion, solving differential equations using the method of undetermined coefficients involves carefully determining the complementary and particular solutions, followed by substituting and simplifying to obtain a complete solution. As a reminder, ensure all coefficients from both sides are correctly equated for an accurate solution!