Solve (ye^xy+1)dx+(xe^xy+2)dy=0,y(0)=0
(ye^xy + 1)dx + (xe^xy + 2)dy = 0
Check for exactness:
Let M = ye^xy + 1
Let N = xe^xy + 2
∂M / ∂y = ∂N / ∂x
∂(ye^xy + 1) / ∂y = ∂(xe^xy + 2) / ∂x
Use chain rule to get the derivative:
e^xy + xy e^xy = e^xy + xy e^xy
Thus it is indeed exact.
To further solve,
∂F/∂x = M = ye^xy + 1
∫ ∂F = ∫ (ye^xy + 1)∂x
F = e^xy + x + g(y)
To get g(y), we differentiate it partially with respect to y:
∂F/∂y = ∂/∂y(e^xy + x + g(y)) = N
∂F/∂y = x e^xy + g'(y) = N
x e^xy + g'(y) = x e^xy + 2
g'(y) = 2
Integrating,
g(y) = 2y
Therefore,
F = e^xy + x + 2y
hope this helps~ `u`
Sorry I forgot to put the constant from integration. Thus,
F = e^xy + x + 2y + C
To solve the differential equation, we will use the method of exact differential equations.
Step 1: Identify if the equation is exact
For the equation (ye^xy+1)dx + (xe^xy+2)dy = 0, we can check if the equation is exact by verifying if the partial derivatives with respect to x and y are equal. Let's find the partial derivatives.
∂/∂x (ye^xy+1) = y(xe^xy+2)
∂/∂y (xe^xy+2) = x(xe^xy+1)
After finding the partial derivatives, we notice that they are not equal. Therefore, the equation is not exact.
Step 2: Make the equation exact
To make the equation exact, we multiply the entire equation by an integrating factor. The integrating factor is given by:
μ(x) = e^∫(∂/∂y - ∂/∂x)dx
In this case, μ(x) will be:
μ(x) = e^∫(x(xe^xy+1) - y(xe^xy+2))dx
Let's calculate the integral:
μ(x) = e^∫(x^2e^xy + xe^xy - yxe^xy - 2ye^xy)dx
= e^((1/3)x^3e^xy - 2xye^xy) + C(y)
So, the integrating factor is μ(x) = e^((1/3)x^3e^xy - 2xye^xy) + C(y), where C(y) is an arbitrary function of y.
Step 3: Multiply the equation by the integrating factor
Multiplying the original equation by the integrating factor μ(x), we get:
e^((1/3)x^3e^xy - 2xye^xy) * (ye^xy+1)dx + e^((1/3)x^3e^xy - 2xye^xy) * (xe^xy+2)dy = 0
Step 4: Simplify and arrange the equation
After multiplying, we simplify and arrange the equation to make it exact.
ye^(2xy)dx + ye^((1/3)x^3e^xy)dx + 2xe^(2xy)dy + 2e^((1/3)x^3e^xy)dy = 0
Step 5: Integrate and solve for C(y)
To solve for C(y), we integrate the equation with respect to y. Since y(0) = 0, C(y) = 0.
∫(ye^(2xy)dx + ye^((1/3)x^3e^xy)dx) + ∫(2xe^(2xy)dy + 2e^((1/3)x^3e^xy)dy) = ∫0dy
Step 6: Solve the integrals
∫(ye^(2xy)dx) = e^(2xy) + g(x), where g(x) is an arbitrary function of x.
∫(ye^((1/3)x^3e^xy)dx) = ∫(ye^u/3 du), where u = x^3e^xy. Using integration by substitution, we have:
Let w = x^3e^xy. Then dw = (3x^2e^xy + x^3e^xy)dy.
Dividing both sides by (3x^2e^xy + x^3e^xy), we get dy = du / (3x^2e^xy + x^3e^xy).
This simplifies the integral to:
∫(ye^((1/3)x^3e^xy)dx) = ∫(ye^u/3 * du) = ∫(1/3) * du = 1/3 * u + g(y) = 1/3 * x^3e^xy + g(y).
Similarly,
∫(2xe^(2xy)dy) = x^2e^(2xy) + h(y), where h(y) is an arbitrary function of y.
∫(2e^((1/3)x^3e^xy)dy) = 2/3 * x^3e^xy + k(x), where k(x) is an arbitrary function of x.
Combining all the terms, we have:
e^(2xy) + 1/3 * x^3e^xy + x^2e^(2xy) + 2/3 * x^3e^xy + g(x) + h(y) + k(x) = C
Simplifying and rearranging, the final solution with the arbitrary functions is:
e^(2xy) + (4/3) * x^3e^xy + g(x) + h(y) + k(x) = C
The given equation is a first-order linear differential equation. To solve it, we'll use an integrating factor. Here's how you can solve it step by step:
Step 1: Rearrange the given equation into the standard form:
(ye^(xy) + 1)dx + (xe^(xy) + 2)dy = 0
Step 2: Notice that the equation is not in the form M(x, y)dx + N(x, y)dy = 0, where M and N are functions of x and y. To convert it to this form, divide both sides of the equation by dx. This gives us:
(ye^(xy) + 1) + (xe^(xy) + 2)(dy/dx) = 0
Step 3: Rearrange the equation to isolate dy/dx as follows:
(ye^(xy) + 1)(dy/dx) = -(xe^(xy) + 2)
Step 4: Now, the equation is in the general form of a first-order linear differential equation, dy/dx + P(x)y = Q(x), where P(x) and Q(x) are functions of x. Here, P(x) = -((ye^(xy) + 1)/(xe^(xy) + 2)) and Q(x) = 0.
Step 5: Calculate the integrating factor (IF). The IF is given by the exponential of the integral of P(x)dx, that is, IF = e^(integral(P(x)dx)). In this case, IF = e^(integral(-((ye^(xy) + 1)/(xe^(xy) + 2))dx)).
Step 6: The integral of P(x)dx can be solved using the substitution method. Let u = xy, then du = ydx + xdy. Rearranging gives us ydx = du - xdy. Substituting this back into the integral, we get:
IF = e^(-(integral((1/(x + 2))du))) = e^(-ln(x + 2)) = 1/(x + 2)
Step 7: Multiply both sides of the rearranged equation (from Step 3) by the integrating factor (IF). We get:
1/(x + 2) * (ye^(xy) + 1)(dy/dx) = 1/(x + 2) * -(xe^(xy) + 2)
Step 8: Simplify the equation by canceling out common terms. This gives us:
y(dy/dx) + (1/(x + 2))(dy/dx) = -1
Step 9: Integrate both sides with respect to x. The left side can be integrated using the substitution method. Integrating y(dy/dx) gives us y. Integating (1/(x + 2))(dy/dx) gives us ln|x + 2|. The right side is simply -x + C, where C is the constant of integration.
Step 10: The equation now becomes:
y + ln|x + 2| = -x + C
Step 11: Apply the initial condition y(0) = 0 to solve for C. Since y(0) = 0, we substitute x = 0 and y = 0 into the equation:
0 + ln|0 + 2| = -0 + C
ln(2) = C
Step 12: Substitute C = ln(2) back into the equation:
y + ln|x + 2| = -x + ln(2)
That's the final solution to the given differential equation, with the initial condition y(0) = 0.