Methods of Solving First Order First Degree Differential Equations

In Mathematics, an equation involving the derivatives of the dependent variable with respect to the independent variable is called the differential equation. The order of the differential equation is defined as the order of the highest order derivative occurring in the differential equation. Similarly, the degree of the differential equation is the highest power of the highest order derivatives that occurs in it. In this article, we are going to discuss the different methods of solving first order first degree differential equations with many solved examples.

Different Methods of Solving First Order First Degree Differential Equations

Now, let us discuss the different methods of solving first order first degree differential equations with solved examples.

Differential Equations with Variable Separable

We know that the first order, first degree differential equation is of the form:

dy/dx = F(x, y) …(1)

If F(x, y) is expressed as the product of g(x) h(y), where g(x) is the function of x and h(y) is the function of y, then the differential equation is said to be of variable separable type. Thus, the differential equation (1) takes the form:

dy/dx = g(x).h(y) …(2)

If h(y) ≠0, and separating the variables, equation (2) becomes

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

Now, integrate the equation (3) on both sides, we get

∫(1/h(y)) dy = ∫g(x) dx …(4)

Hence, equation (4) provides the solution for the differential equation in the form:

H(y) = G(x)+C

Where H(y) and G(x) are the antiderivatives of 1/h(y) and g(x), respectively and C is called the arbitrary constant.

Example 1:

Find the particular solution of the differential equation dy/dx = -4xy² given that y=1 and x = 0.

Solution:

Given differential equation: dy/dx = -4xy²

Here, y≠0, and hence the given differential equation is written as:

dy/y² = -4x dx (Using Variable separable method) …(1)

Now, integrate the equation (1) on both sides, we get

∫dy/y² = -4 ∫x dx

-1/y = -2x²+C

Therefore,

y= 1/(2x²-C) …(2)

Substituting x = 0 and y= 1 in equation (2), we get

1 = 1/(2(0)²-C)

1 = 1/-C

Hence, C = -1

Now, substitute C = -1 in equation, we get

y =1/(2x²+1), which is the particular solution of the given differential equation dy/dx = -4xy².

Homogeneous Differential Equations

A differential equation of the form dy/dx = F(x, y) is called a homogeneous differential equation if F(x, y) is a homogeneous function of degree zero.

Hence, to solve the homogeneous differential equation of the type, dy/dx = F(x, y) = g(y/x) …(1)

we should make the substitution, y = v.x … (2)

Now, differentiate the equation (2) with respect to x, we get

dy/dx = v+ x(dv/dx) …(3)

Now, substituting the value of dy/dx from equation (3) in equation (1), we get

v+x(dv/dx) = g(v)

The above equation can also be written as:

x (dv/dx) = g(v)-v ..(4)

Now, separate the variables from equation (4), we get

Now, integrate the equation (5) on both sides, we get

∫ [dv/(g(v)-v] = ∫dx/x + C …(6)

Therefore, equation (6) gives the primitive solution of the given differential equation if we replace v by y/x.

Example 2:

Show that the differential equations xcos(y/x)(dy/dx) = y cos(y/x) + x is homogeneous and solve it.

Solution:

Given differential equation: xcos(y/x)(dy/dx) = y cos(y/x) + x

The given equation is also written as:

dy/dx = [ycos(y/x)+x] / [x cos (y/x)] …(1)

The above equation is a differential equation of the form dy/dx = F(x, y)

Here, F(x, y) = [ycos(y/x)+x] / [x cos (y/x)]

Now, replace x by λx and y by λy, we get

F(λx, λy) = [λ[ycos(y/x)+x]] / [λ[x cos (y/x)]]

F(λx, λy) = λ⁰ [F(x, y)]

Hence, F(x, y) is a homogeneous function of degree zero.

So, the differential equation is a homogeneous differential equation, and hence we make the substitution, y = v.x …(2)

Differentiate the equation (2) with respect to the variable x, we get

(dy/dx) = v+ x(dy/dx) …(3)

Now, substitute the value of y and dy/dx in equation (1),

v + x(dv/dx) = (v cos v + 1)/(cos v)

⇒ x(dv/dx) = [(v cos v + 1)/(cos v)] – V

⇒ x (dv/dx) = 1/cos v

⇒ cos v dv = dx/x

Now, integrate the equation on both sides,

∫ cos v dv = ∫(1/x) dx

⇒ sin v = log |x| + log |C|

⇒ sin v = log |Cx|

Replace v by y/x, we get

Sin (y/x) = log |Cx|, which is the general solution for the given differential equation xcos(y/x)(dy/dx) = y cos(y/x) + x.

Linear Differential Equations

If the differential equation is of the form, (dy/dx) + Py = Q, then it is known as a first-order linear differential equation.

Here,

P and Q are constants or the functions of x.

Follow the below procedure to solve the first-order linear differential equation of the type

(dy/dx) + Py = Q …(1)

Multiply both sides of equation by the function of x, say g(x),

g(x). (dy/dx) + P.g(x). y = Q. g(x) …(2)

Choose g(x) in such a way that the right hand side of the equation becomes the derivative of y. g(x).

Thus, equation (2) becomes:

g(x) (dy/dx) + P. g(x). y = (d/dx) [y. g(x)]

⇒ g(x) (dy/dx) + P.g(x)y = g(x)(dy/dx) + y g’(x)

⇒ P. g(x) = g’(x)

⇒ P = g’(x)/g(x).

Now, integrate both sides of equation with respect to x, we get

∫ P dx = ∫ g’(x)/g(x) dx.

⇒ ∫P.dx = log (g(x))

⇒ g(x) = e^{∫P. dx}.

Multiply the equation (1) by g(x) = e^{∫P. dx}. Thus, the left-hand side of the derivative will be some functions of x and y. Hence, the function g(x) = e^{∫P. dx} is called the integrating factor for the given differential equation.

On substituting the value of g(x) in equation (2), we get,

e^{∫P. dx}. (dy/dx) + P.e^{∫P. dx}. y = Q. e^{∫P. dx} (or)

⇒ (d/dx) ( ye^{∫P. dx}) = Qe^{∫P. dx}.

Now, integrate the above equation with respect to the variable x, we get

y.e^{∫P. dx} = ∫(Q.e^{∫P. dx})dx

y = e^{-∫P. dx}. ∫(Q.e^{∫P. dx})dx + C, which is called the general solution for the given differential equation.

Summary:

1. Write the differential equation in the form (dy/dx)+Py= Q, where P and Q are the functions of x.
2. Determine the integrating factor, I. F = e^{∫P. dx}.
3. Write the solution for the given differential equation in the form:

y.(I.F) = ∫ (Q (I.F) dx + C

Example 3:

Find the solution for the differential equation x (dy/dx) + 2y = x², where x≠0.

Solution:

Given differential equation: x (dy/dx) + 2y = x² …(1)

Now, divide both sides of equation (1) by x, we get

(dy/dx) + 2(y/x) = x , which is the linear differential equation of the form (dy/dx) + Py = Q.

Here, P = 2/x and Q = x

Therefore, the Integrating factor, I.F = e^{∫(2/x). dx}

I.F = e^{2 log x}

I.F = e ^{log x power 2}

I.F = x² (Because e^{log f(x)} =f(x))

Therefore, the solution for the given linear differential equation is:

y . x² = ∫(x) (x²)dx + C = ∫ x³ dx + C

Therefore, y = (x²/4) + Cx^-2 is the general solution for the given differential equation x (dy/dx) + 2y = x^2.

Practice Problems

1. Find the general solution for the differential equation (dy/dx) + (y/x) = x².
2. Show that the differential equation (x-y) dy – (x+y)dx = 0 is homogeneous and solve it.
3. Find the general solution of the differential equation (dy/dx) + y = 1, where y≠1.
4. The principal amount increases continuously at the rate of r% per year in a bank. Determine the value of r, if Rs, 100 doubles itself in 10 years. (Use: log_e 2 = 0.6931).

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