In Exercises 1–22, solve the differential equation.

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Question

In Exercises 1–22, solve the differential equation.

dy + x(2y - e^(x-x²))dx = 0

Accepted Answer

Welcome back everyone to another video. Solve the differential equation dyy minus x multiplied by 4 y plus 3 e to the power of x 2 dx equals 0. So for this problem we are going to rewrite the equation in its standard form. What we're going to do in the first step is divide both sides by dx and distribute. So we're going to get the dyy divided by dx, which gives us the derivative. -4 xy minus 3 xe to the power of x 2 is equal to 0. And now we are going to move the function of x to the right hand side. Specifically, we are going to get the d y divided by d x minus 4 x y equals 3 x e to the power of x squared. From here we will find the integrating factor. Let's remember how we find it, specifically mu, which is the integrating factor, is e to the power of the integral of -4 x, right, which is the term in front of y. So -4 XDX. Let's go ahead and integrate. That's E. Raise to the power of -4 multiplied by x2 divided by 2, which gives us -2 x 2. Now, our next step is to multiply both sides of the equation by the integrating factor. Which gives us e to the power of -2 x 2, DY divided by DX. -4 x e to the power of -2 x 2 is equal to 3 x. And now we get each the power of x 2 multiplied by. Eats the power of -2 x 2. If we add the exponents, we are going to get x2. So overall. We will get 3 x e to the power of x 2. On the left hand side, we have the derivative of a product, so that's d divided by dx of y multiplied by e to the power of -2 x 2. On the right hand side we have 3 x e to the power of x squared and what we're going to do from here is multiply both sides by dx so that d of ye to the power of -2 x squared is equal to 3 x e to the power of x 2 d x. And now we can integrate both sides. Let's begin with the integral on the right-hand side, and we're going to use the U substitution. Let's suppose that U is X2. DU divided by dx is the derivative of x 2 which is -2 x. And now let's go ahead and identify xDx because that's what we have within the integral. That would be DU divided by -2 or 1/2 DU. Now we will rewrite the integral. We have 3 integral xe to the power of x 2 d x. We have factored out the constant, right, so we get 3 integral. It's the power of you. Multiplied by 1/2 DU, so we can just write DU and factor out -1/2. And this integral is going to be equal to -3 halves e to the power of u plus a constant of integration c. And since u is x2, we get -3 halves e to the power of x2 plus a constant of integration c. Now if we go back into our equation in terms of the two integrals on the left hand side, we have an integral of the differential, so we get the differential function itself, which is ye to the power of -2 x 2, and on the right hand side we already know that we have -3 halves. E to the power of x2 + C, right? Well done. So now from here we will identify. The solution for y. What we are going to do is multiply both sides by e to the power of 2 x 2, which gives us y equals. -3 halves eats the power off. X2 + 2 X2 gives us X2, right? So this is the exponent plus c multiplied by e to the power of 2 x 2. And this is our final solution for this problem. Thank you for watching.

In Exercises 1–22, solve the differential equation.dy + x(2y - e^(x-x²))dx = 0

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In Exercises 1–22, solve the differential equation.

dy + x(2y - e^(x-x²))dx = 0