Solution Of Differential Equations

We begin by addressing the differential equation: . This is a first-order linear differential equation in standard form , where and .

To solve this, we find an integrating factor . Here, . Thus, the integrating factor is .

Multiplying the entire differential equation by this integrating factor gives:

.

This simplifies to:

.

The left side is now the derivative of the product :

.

Integrating both sides with respect to gives:

, where is the integration constant.

Solving for , we multiply through by :

.

This is the general solution to the differential equation.

Thus, the correct answer is , where is a constant.

The given differential equation is . To find the general solution, we start by recognizing this as a first-order homogeneous differential equation. The equation can be rewritten as:

This is a separable differential equation. We can separate the variables by rearranging terms to get:

Integrate both sides:

Upon integration, we obtain:

We can rewrite the right side by factoring as follows:

This simplifies to:

Exponentiating both sides to remove the natural logarithm gives:

Let , a positive constant, so we have:

This implies the general solution to the differential equation is:

The given differential equation is . This is a separable differential equation, which means we can rearrange the terms to separate variables and . Let's proceed with the solution.

First, separate the variables:

Integrate both sides:

This yields:

, where is the constant of integration.

Multiplying the entire equation by 2 to eliminate fractions:

Rearranging gives:

This matches the correct solution provided: , where is a constant of integration.

To find the number of solutions for the differential equation with the initial condition , we proceed by checking the type of differential equation and solving it. First, rewrite the equation:

Rearranging terms, we have:

This can be rewritten as:

The equation is separable, as we can express it as:

The integrals can be computed by appropriate substitutions. For simplicity, assume , substituting and simplifying leads to a general solution. Then, apply the given condition to find the constant:

This yields:

With the initial condition , we find:

Substituting back gives the unique exact solution:

This calculation shows that there is only one solution, as any solution would have to satisfy the given initial condition.

To solve the differential equation cosx siny dx + sinx cosy dy = 0, we can use the method of exact equations. An equation of the form M(x,y)dx + N(x,y)dy = 0 is exact if there exists a function ψ(x,y) such that:

∂ψ/∂x = M and ∂ψ/∂y = N.

Given:

M(x,y) = cosx siny

N(x,y) = sinx cosy

We check if the equation is exact by verifying:

∂M/∂y = ∂(cosx siny)/∂y = cosx cosy

∂N/∂x = ∂(sinx cosy)/∂x = cosx cosy

Since ∂M/∂y = ∂N/∂x, the equation is exact.

Now, we find the function ψ(x,y):

∫M(x,y)dx = ∫cosx siny dx = siny ∫cosx dx = siny sinx + h(y)

Next, we differentiate ψ(x,y) with respect to y:

∂ψ/∂y = sinx cosy + h'(y)

Equating it to N(x,y), we get:

sinx cosy = sinx cosy + h'(y)

This implies h'(y) = 0 which means h(y) is a constant, denoted as C.

Hence, the solution is ψ(x,y) = sinx siny = C.

The correct solution is: sinx siny = C, where C is a constant.

To solve the differential equation with the initial condition , proceed as follows:

First, separate the variables to integrate:

Integrating both sides, we get:

Next, use the initial condition to find the constant :

Simplifying, we have:

Thus, .

The particular solution is:

The correct answer is .

To solve the differential equation , we start by rearranging it into a form suitable for integration. Rewrite the equation as:

This is a first-order linear differential equation in the form , where and . To solve it, we need to determine the integrating factor, which is .

Calculating the integrating factor:

Then, .

Multiplying both sides of the differential equation by the integrating factor:

Now, the left-hand side is the derivative of with respect to . Integrate both sides with respect to :

Integrating gives:

Assume solution in the form . Testing this satisfies the original differential equation.

Hence, the correct solution is: