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Solve the deffrintal equation

3) x y'=2 y+x^{3} e^{x}

User Zenab
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Answer:

Explanation:

To solve the given differential equation, which is a first-order linear ordinary differential equation (ODE), we can use the method of separation of variables.

The given differential equation is: xy' = 2y + x^3 * e^x

First, we'll separate the variables y and x by moving all y-related terms to one side and all x-related terms to the other side:

xy' - 2y = x^3 * e^x

Next, we'll divide both sides by x to make the equation look like the standard form of a first-order linear ODE:

y' - (2/x)y = x^2 * e^x

Now, we'll find the integrating factor (IF) to solve the equation. The integrating factor is given by the exponential of the integral of the coefficient of y:

IF = e^(∫(-2/x) dx)

IF = e^(-2 ln|x|)

IF = e^(ln|x^(-2)|)

IF = |x^(-2)|

Now, we'll multiply both sides of the equation by the integrating factor:

|x^(-2)| * y' - 2|x^(-2)| * y = |x^(-2)| * x^2 * e^x

The left side can be written as the derivative of the product of the integrating factor and y:

d/dx (|x^(-2)| * y) = |x^(-2)| * x^2 * e^x

Now, we'll integrate both sides with respect to x:

∫ d/dx (|x^(-2)| * y) dx = ∫ |x^(-2)| * x^2 * e^x dx

The left side is just the integral of a derivative, so it simplifies to:

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

Now, we need to evaluate the integral on the right side. To do this, we'll consider two cases: x > 0 and x < 0.

Case 1: x > 0

∫ x^2 * e^x dx

We can integrate this by parts:

Let u = x^2 (differential: du = 2x dx)

Let dv = e^x dx (differential: v = e^x)

∫ x^2 * e^x dx = x^2 * e^x - ∫ 2x * e^x dx

∫ x^2 * e^x dx = x^2 * e^x - 2 ∫ x * e^x dx

Now, we'll integrate the remaining integral on the right side by parts again:

Let u = x (differential: du = dx)

Let dv = e^x dx (differential: v = e^x)

∫ x * e^x dx = x * e^x - ∫ 1 * e^x dx

∫ x * e^x dx = x * e^x - e^x

Now, substituting this back into the previous result:

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

∫ x^2 * e^x dx = x^2 * e^x - 2x * e^x + 2e^x

Case 2: x < 0

In this case, the absolute value |x^(-2)| is just 1/x^2, and we integrate in a similar way:

∫ x^2 * e^x dx = x^2 * e^x + 2x * e^x + 2e^x

Now that we have evaluated the integral for both cases, we can continue with the solution:

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

|x^(-2)| * y = {

x^2 * e^x - 2x * e^x + 2e^x, if x > 0,

x^2 * e^x + 2x * e^x + 2e^x, if x < 0

}

We can simplify the right side by removing the absolute value, as the function x^(-2) is positive for both positive and negative x:

y = {

x^2 * e^x / x^2 - 2x * e^x / x^2 + 2e^x / x^2, if x > 0,

x^2 * e^x / x^2 + 2x * e^x / x^2 + 2e^x / x^2, if x < 0

}

Now, simplify further:

y = {

e^x - 2x * e^x / x^2 + 2e^x / x^2, if x > 0,

e^x + 2x * e^x / x^2 + 2e^x / x^2, if x < 0

}

Finally, we can express the solution in a unified form:

y = e^x + 2x * e^x / x^2 + 2e^x / x^2

So, the solution to the given differential equation is y = e^x + 2x * e^x / x^2 + 2e^x / x^2.

User Dty
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