Solve dy dx − 2xy = x, with y(0) = 0. Purchase the answer t…

To solve the differential equation dy/dx – 2xy = x, with the initial condition y(0) = 0, we will use the method of integrating factors. This method involves multiplying the entire equation by a certain function, known as the integrating factor, in order to make the left side of the equation a derivative of a product. This will allow us to easily integrate both sides and find the general solution.

First, let’s rewrite the equation in the standard form:
dy/dx = 2xy + x.

Now, let us focus on finding the integrating factor. The integrating factor is given by the formula:

μ(x) = e^(∫(-2x) dx) = e^(-x^2).

Multiplying both sides of the equation by the integrating factor, we get:

e^(-x^2) * dy/dx = e^(-x^2) * (2xy + x).

Now, notice that the left side of the equation is the derivative of (e^(-x^2) * y) with respect to x. So, we can rewrite the equation as:

d/dx (e^(-x^2) * y) = e^(-x^2) * (2xy + x).

To integrate both sides, we get:

∫d/dx (e^(-x^2) * y) dx = ∫e^(-x^2) * (2xy + x) dx.

Simplifying the left side using the Fundamental Theorem of Calculus and applying the chain rule, we have:

e^(-x^2) * y = ∫2x * e^(-x^2) * y dx + ∫x * e^(-x^2) dx.

Now, let’s solve each integral separately. Firstly, we can factor out the terms of y from the first integral:

∫2x * e^(-x^2) * y dx = 2y * ∫x * e^(-x^2) dx.

The integral on the right side can be evaluated using a substitution. Let u = -x^2, so du = -2x dx. After the substitution, we have:

∫x * e^(-x^2) dx = -∫e^u du = -e^u + C = -e^(-x^2) + C_1,

where C is the constant of integration.

Substituting back into the equation, we get:

e^(-x^2) * y = 2y * (-e^(-x^2) + C_1) + (-e^(-x^2) + C_2).

Combining like terms and simplifying, we have:

e^(-x^2) * y = -2ye^(-x^2) + 2C_1y – e^(-x^2) + C_2.

To find the constant C_2, we use the initial condition y(0) = 0. Substituting x = 0 and y = 0 into the equation, we get:

e^(-0^2) * 0 = -2(0)e^(-0^2) + 2C_1(0) – e^(-0^2) + C_2.

Simplifying this expression, we find:

C_2 = 0.

Substituting this back into the equation, we have:

e^(-x^2) * y = -2ye^(-x^2) + 2C_1y – e^(-x^2).

Rearranging the equation, we get:

y(e^(-x^2) + 2e^(-x^2)) = -e^(-x^2) + 2C_1y.

Now, dividing both sides of the equation by (e^(-x^2) + 2e^(-x^2)), we have:

y = (-e^(-x^2) + 2C_1y) / (e^(-x^2) + 2e^(-x^2)).

Simplifying further, we get:

y = (-e^(-x^2) + 2C_1y) / 3e^(-x^2).

Now, let’s find the constant C_1. To do this, we use the initial condition y(0) = 0. Substituting x = 0 and y = 0 into the equation, we get:

0 = (-e^(-0^2) + 2C_1(0)) / 3e^(-0^2).

Simplifying this expression and solving for C_1, we find:

C_1 = -1/2.

Substituting this back into the equation, we have:

y = (-e^(-x^2) – y/2) / 3e^(-x^2).

Further simplifying the equation, we get:

3ye^(-x^2) = -e^(-x^2) – y/2.

Combining like terms, we have:

3ye^(-x^2) + y/2 = -e^(-x^2).

Now, we can solve for y in terms of x:

y = (-e^(-x^2) – y/2) / 3e^(-x^2).

Multiplying both sides of the equation by 3e^(-x^2), we get:

3ye^(-x^2) = -e^(-x^2) – y/2.

Now, let’s simplify the right side of the equation:

3ye^(-x^2) = -1/2e^(-x^2) – y/2.

Combining like terms, we have:

3ye^(-x^2) + y/2 = -1/2e^(-x^2).

Multiplying both sides of the equation by 2, we get:

6ye^(-x^2) + y = -e^(-x^2).

Finally, we can solve for y:

y = -1/(6e^(-x^2)).

Therefore, the solution to the given differential equation dy/dx – 2xy = x, with the initial condition y(0) = 0, is y = -1/(6e^(-x^2)).