# Solutions to Ordinary Differential Equations of First Order

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There are so many aspects a person has to keep in mind while solving differential equations that you’re bound to forget something at some point. This can be very frustrating as by making a small mistake, differential equations can become unsolvable. The purpose for writing this is to create a helping guide to help remember everything.

# Variable Separable ODEs

I’ll start with variable separable differential equations. Although a variable separable ODE should be in the form

dy/dx = g(x)h(y)

which you can easily solve by applying algebraic operations and converting it into the form

h(y)dy = g(x)dx

and then finally integrating both sides to obtain the solution. This is rarely the case and the ODE we generally deal with are not provided to us in such a simple form. To solve these equations, just apply algebraic operations to separate the variables and bring it into the form mentioned above and integrating it.

# Linear Differential Equations

There are two types of linear equations, homogenous and non-homogenous differential equations. These equations are generally in the form

a(x) dy/dx + b(x)y = g(x).

The reason why these equations are called linear is because the dependent variable(y) has a power of one. To solve them, we first have to bring them in the form

dy/dx + P(X)y = f(x)

If f(x) in this form comes out to be zero, it’s called a homogenous differential equation, otherwise it is a non-homogenous differential equation.

To solve these, we first find the integrating factor. We do this by integrating P(x) and setting it as the power of *e. *We then multiply the integrating factor[IF] with the entire equation giving us the equation in the form

[IF]dy/dx + [IF]P(x)y = [IF]f(x)

which is equivalent to

d/dx [IF] . y = [IF]f(x)

Integrating both sides should give us the solution.

# Exact Equations

The criteria for an equation to be exact is that if we write it in the form

M(x,y)dx + N(x,y)dy = 0

the partial derivative of M with respect to y and the partial derivative of N with respect to x should be continuous that is

∂M/∂y = ∂N/∂x

Once it is established that the equation is exact, for the solution we apply integration in the form

∫ M dx + ∫ (Terms of N free from x) dy = C

or

∫ N dy + ∫ (Terms of M free from y) dx = C

after integration we can obtain the solution for the equation.

Even if a differential equation is non-exact, we can make it exact by finding its integrating factor and multiplying with it. We find its integrating factor by taking the partial derivatives and putting them in the formula

P(x) = (∂M/∂y - ∂N/∂x)/ N

or

P(x) = (∂N/∂x - ∂M/∂y)/ M

we then integrate P(x) and set it as the power of *e. *We then multiply both M and N by the integrating factor to make the differential equation exact and then test it for exactness again by finding the partial derivatives.

# Homogenous Differential Equations

To solve a homogenous differential equation, we first have to perform a homogeneity test. We do this by solving the equation by putting the values of x and y as λx and λy respectively. If the equation comes out in the form where

M(λx, λy) = λªM(x, y)

and

N(λx, λy) = λⁿN(x, y)

where n is the degree of homogeneity. Once it is verified that the given equation is a homogenous differential equation, we come towards solving it. To solve it we perform a substitution of either y = ux or x = uy. We then take the derivative of this substitution and write it in the form

dy = u dx + x du

if we take the example of y = ux. We then perform the substitutions to bring the equation in the form where the variables are u and x. We then perform the integration to obtain our solution in the form of u and x. We resubstitute the value of u to obtain our final solution.

# Bernoulli’s Equations

The differential equations in the form

dy/dx + P(x)y = Q(x)yⁿ

where n is zero, is called a Bernoulli differential equation. If n = 1, it can also be said to be a Bernoulli differential equation as it can be written in the form

dy/dx + [P(x)y-Q(x)yⁿ] = 0

To solve Bernoulli equations, we first divide the entire equation by yⁿ to obtain

y ⁻ⁿ dy/dx P(x) y ⁻ⁿ = Q(x)

and then make the substitution where

u = y¹⁻ⁿ

By using chain rule we get

du/dx = (1 - n)y⁻ⁿ. dy/dx

which can also be written as

y ⁻ⁿ. dy/dx = du/dx . 1/(1 - n)

After substitution we get

du/dx . 1/(1 - n) + P(x)u = Q(x)

By algebraic manipulation we get

du/dx + (1 - n)P(x)u = (1 - n)Q(x)

which can be solved as a linear differential equation linear in u. Once a solution is obtained in u, u can be re-substituted to obtain a solution in y.