integrate it by integrating factor
(cos^3x)dy/dx +ycosx=sinx
Hmmm. rearrange things a bit to get
y' + sec^2 x y = tan x sec^2 x
that is
y' + P(x)y = Q(x)
integrating factor is thus e^∫sec^2 x dx = e^(tan x)
now plug and chug
To integrate the given differential equation using an integrating factor, follow these steps:
Step 1: Identify the standard form of the linear first-order ordinary differential equation (ODE):
(dy/dx) + P(x)y = Q(x)
In this case, the given equation is:
(dy/dx) + ycosx = sinx
Step 2: Determine the values of P(x) and Q(x) in the standard form of the ODE.
P(x) = cosx
Q(x) = sinx
Step 3: Calculate the integrating factor (IF) using the formula:
IF = exp(∫ P(x) dx)
In this case,
IF = exp(∫ cosx dx)
Step 4: Integrate the expression ∫ P(x) dx to find the integrating factor.
∫ cosx dx = sinx
Therefore,
IF = exp(sinx) = e^(sinx)
Step 5: Multiply the entire equation by the integrating factor (IF):
e^(sinx) * (dy/dx) + e^(sinx) * ycosx = e^(sinx) * sinx
The left-hand side of the equation can be simplified using the product rule:
(d/dx)(e^(sinx) * y) = e^(sinx) * sinx
Step 6: Integrate both sides of the equation:
∫ (d/dx)(e^(sinx) * y) dx = ∫ e^(sinx) * sinx dx
Using the fundamental theorem of calculus, we can perform these integrations:
e^(sinx) * y = -e^(sinx) * cosx + C
Step 7: Solve for y:
y = -cosx + Ce^(-sinx)
Where C is the arbitrary constant that appears during integration.
To integrate the given differential equation using the integrating factor method, follow these steps:
Step 1: Identify the form of the differential equation.
The given equation is a first-order linear ordinary differential equation. It can be written in the form:
dy/dx + P(x)y = Q(x),
where P(x) = ycos(x) and Q(x) = sin(x).
Step 2: Find the integrating factor.
The integrating factor (IF) is given by the formula:
IF = e^(∫P(x)dx).
In this case, P(x) = ycos(x), so we need to evaluate the integral of ycos(x) with respect to x.
Step 3: Solve the integral to find the integrating factor.
To evaluate the integral, we treat 'y' as a constant with respect to integration. So ∫ycos(x)dx = y∫cos(x)dx.
∫cos(x)dx = sin(x) + C₁, where C₁ is the constant of integration.
Therefore, the integrating factor (IF) = e^(∫P(x)dx) = e^(y∫cos(x)dx) = e^(ysin(x) + C₁) = e^(ysin(x)) * e^(C₁).
Step 4: Multiply the entire differential equation by the integrating factor.
Multiply both sides of the differential equation by the integrating factor (e^(ysin(x)) * e^(C₁)):
e^(ysin(x)) * e^(C₁) * (cos^3(x))(dy/dx) + e^(ysin(x)) * e^(C₁) * ycos(x) = e^(ysin(x)) * e^(C₁) * sin(x).
Step 5: Simplify the left side of the equation.
Notice that e^(C₁) is a constant, so we can combine it with the integrating factor. Let's denote e^(C₁) as 'K' for simplicity.
K * (cos^3(x))(dy/dx) + Kycos(x) = Ksin(x).
Step 6: Rewrite the left side of the equation as a derivative.
Let's take the derivative of both sides with respect to 'x'.
(d/dx)[K * (cos^3(x))(y)] = Ksin(x).
Using the product rule, the left side can be expanded as:
K * (cos^3(x))(dy/dx) + [3K(cos^2(x))(sin(x))]y.
Step 7: Simplify the equation.
The equation now becomes:
[d/dx)[K * (cos^3(x))(y)] = Ksin(x) - [3K(cos^2(x))(sin(x))]y.
Step 8: Integrate both sides of the equation.
Integrate both sides with respect to 'x'.
∫[d/dx){K * (cos^3(x))(y)] dx = ∫[Ksin(x) - 3K(cos^2(x))(sin(x))] dx.
Integration on the left side gives: K * (cos^3(x))(y).
Integration on the right side gives: -Kcos(x) - Kcos^3(x)/3 + C₂, where C₂ is the constant of integration.
Step 9: Solve for 'y'.
We now have: K * (cos^3(x))(y) = -Kcos(x) - Kcos^3(x)/3 + C₂.
Dividing both sides by K * (cos^3(x)):
y = (-cos(x) - cos^3(x)/3 + C₂) / (cos^3(x)).
And that's the solution to the given differential equation using the integrating factor method.