A differential equation is an equation which involves derivatives of a dependent variable with respect to other independent variables. It is a powerful tool used in a variety of fields such as Mathematics, medical, chemistry, physics, and engineering to build mathematical models. In this section, we will explore differential equations in detail, along with solved examples.

Differential equations are equations that involve derivatives of a function with respect to one or more of its variables. They are used to describe the behavior of a wide variety of physical systems, such as electrical circuits, mechanical systems, and biological systems.

A differential equation is an equation involving a function and one or more of its derivatives. It is also defined as an equation that contains derivatives of one or more dependent variables with respect to one or more independent variables. If a function has only one independent variable, then it is an ordinary differential equation.

Examples of Differential Equations:

• Separable Differential Equations
• Exact Differential Equations
• Linear Differential Equations
• Bernoulli Differential Equations
• Homogeneous Differential Equations
• Nonlinear Differential Equations

1. (\frac{d^2y}{dx^2} - \frac{dy}{dx} - 4x = 0)

2. $\left( \frac{d^2y}{dx^2} \right) - 6\frac{dy}{dx} + 8y = 0$

3. $$\left[ 2+\left( \frac{dy}{dx} \right) \right]={{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{\frac{2}{3}}}$$

Note: A Partial Differential Equation is an equation involving a function of several variables and its partial derivatives.

Degree and Order of Differential Equations

Order: Highest Differential Coefficient

Degree: The highest power of the derivatives in a differential equation expressed as a Polynomial is known as the Degree.

(\begin{array}{l}f\left( xy \right)\frac{{d}^{m}}{d{{x}^{m}}}y+g\left( xm \right)\frac{{d}^{m-1}}{d{{x}^{m-1}}}y+…=0\end{array} )

Reminder:

The degree of a differential equation is only defined when it is in polynomial form.

Different Ways to Define a Degree:

* (y’ + 6y^2 + y = 0 \ \text{at} \ (1, 1))

* (y’ + y - \sqrt{1} = 0 \ \text{at} \ (1,2))

* (y + \left(2y\right)^6 + \sin(y) = 0 \ \text{at} \ (2, 1))

* $y’+\sin(y)+y=0$ at $(2, ND)$

Formation of Differential Equations

Let $y$ and $x$ be the dependent and independent variables, respectively, for the equation where $k$ is an arbitrary constant. Now, here is the step to form a differential equation:

Count the number of independent arbitrary constants (let $n$ be the number of constants)

The order of the differential equation will be equal to the number of arbitrary constants. (n)

Differentiate the given equation ’n’ times to eliminate arbitrary constants.

The above elimination equation will be a required differential equation.

For family of curves: (\begin{array}{l}f\left( x,y,\alpha ,{{\alpha }{2}}….{{\alpha }{n}} \right) = 0 \end{array})

If we differentiate $n$ times, we get the differential equation of equation (1), where $\alpha_1$, $\alpha_2$, $\dots$, $\alpha_n$ are $n$ different parameters.

Answer:

Find the equation of all lines passing through the origin in the form of a differential equation.

This is a bold statement.

Answer: This is a bold statement.

‘\(y = mx\)’

‘(\frac{dy}{dx} = m)’

Therefore, xdy - ydx = 0 is the answer.

#Solution of Differential Equations

The relation between the variables involved which satisfies the differential equation is the solution of the differential equation.

Types of Solutions:

1. General Solution:

It contains arbitrary constants equal to the order of the differential equation.

2. Particular Solution

The general solution of the differential equation is obtained by assigning values to the arbitrary constants.

Five Methods for Solving a Differential Equation

1. Solution by Inspection

2. Variable Separable

3. Homogeneous

4. Linear Differential Equation

5. General

Solution by Inspection

If the differential equation is of the form $$f(f_1(x,,y))\ d(f_1(x,,y))+\varphi (f_2(x,,y))\ d(f_2(x,,y))+\dots=0,$$ then each term can be separately integrated.

The inspection method can be used to find the solution to a differential equation. Additionally, it is important to memorize the associated results.

(i) $\mathrm{d}(x+y) = \mathrm{dx} + \mathrm{dy}$

(ii) $\mathrm{d}(xy) = x \mathrm{dy} + y \mathrm{dx}$

(iii) $\mathrm{d}\left(\frac{x}{y}\right) = \frac{y \mathrm{dx} - x \mathrm{dy}}{y^{2}}$

(iv) $\mathrm{d}\left(\frac{y}{x}\right) = \frac{x \mathrm{dy} - y \mathrm{dx}}{x^{2}}$

(v) $\mathrm{d}\left(\frac{x^{2}}{y}\right) = \frac{2xy \mathrm{dx} - x^{2} \mathrm{dy}}{y^{2}}$

(vi) $\mathrm{d}\left(\frac{y^{2}}{x}\right) = \frac{2x \mathrm{dy} - y^{2} \mathrm{dx}}{x^{2}}$

(vii) $\mathrm{d}\left(\frac{x^{2}}{y^{2}}\right) = \frac{2xy^{2} \mathrm{dx} - 2x^{2} \mathrm{dy}}{y^{4}}$

(viii) $\mathrm{d}\left(\frac{y^{2}}{x^{2}}\right) = \frac{2x^{2} \mathrm{dy} - 2xy^{2} \mathrm{dx}}{x^{4}}$

(ix) $\mathrm{d}\left(\tan^{-1}\frac{x}{y}\right) = \frac{y \mathrm{dx} - x \mathrm{dy}}{x^{2} + y^{2}}$

(x) $\mathrm{d}\left(\tan^{-1}\frac{y}{x}\right) = \frac{x \mathrm{dy} - y \mathrm{dx}}{x^{2} + y^{2}}$

(xi) $\mathrm{d}\left(\ln[xy]\right) = \frac{x \mathrm{dx} + y \mathrm{dx}}{xy}$

(xii) $\mathrm{d}\left(\frac{x}{y}\right) = \frac{y \mathrm{dx} - x \mathrm{dy}}{xy}$

(xiii) $\mathrm{d}\left(\frac{1}{2}\ln(x^{2} + y^{2})\right) = \frac{x \mathrm{dx} + y \mathrm{dy}}{x^{2} + y^{2}}$

(xiv) $\mathrm{d}\left[\ln \frac{y}{x}\right] = \frac{x \mathrm{dy} - y \mathrm{dx}}{xy}$

(xv) $\mathrm{d}\left(\frac{-1}{xy}\right) = \frac{x \mathrm{dy} + y \mathrm{dx}}{x^{2}y^{2}}$

(xvi) $\mathrm{d}\left(\frac{e^{x}}{y}\right) = \frac{ye^{x} \mathrm{dx} - e^{x} \mathrm{dy}}{y^{2}}$

(xvii) $\mathrm{d}\left(\frac{e^{y}}{x}\right) = \frac{x e^{y} \mathrm{dy} - e^{y} \mathrm{dx}}{x^{2}}$

(xviii) $\mathrm{d}\left(x^{m}y^{n}\right) = x^{m-1}y^{n-1}(m \mathrm{y} \mathrm{dx} + nx \mathrm{dy})$

(xix) $\mathrm{d}\left(\sqrt{x^{2} + y^{2}}\right) = \frac{x \mathrm{dy} - y \mathrm{dx}}{x^{2} - y^{2}}$

**(xx

Variable Separable Method

If an equation can be written with variables separated for integration, then it can be solved.

(\int{f\left( x \right) , dx + \int{g\left( y \right) , dy = c})

Homogeneous Differential Equation

A differential equation of form (\frac{dy}{dx}=\frac{f\left( x,y \right)}{\phi \left( x,y \right)}) where (f) and (\phi) are homogeneous is called a homogenous differential equation.

Homogenous Function: The function f(x, y) is called a homogenous function if f(ax, ay) = a^nf(x, y), where a is a constant and n is a constant.

(\begin{array}{l}f\left( \lambda x, \lambda y \right) = \lambda^n \cdot f\left( x,y \right)\end{array})

Thus, a homogeneous function can be written as

(\begin{array}{l}f\left( x,y \right)={{x}^{n}}f\left( \frac{y}{x} \right)\end{array} )

or

(\begin{array}{l}f\left( x,y \right)={{y}^{n}}f\left( \frac{y}{x} \right)\end{array} )

Homogeneous Differential Equations

In first-order, first-degree differential equations are expressed in the form of: $$\frac{dy}{dx} = f(x,y)$$

(\frac{dy}{dx}=\frac{f\left( x,y \right)}{g\left( x,y \right)})

**Solution:**The solution to the differential equation $x^2 \frac{dy}{dx} + y(x + y)\frac{dx}{dy} = 0$ with the initial condition $y = 1$ when $x = 1$ is $y = x^2 + 1$.

Given:

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Solution:

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\(\begin{array}{l}{{x}^{2}}\ dy + y\left(x + y\right)\ dx = 0\end{array}\)

(\begin{array}{l}{-y\left( x+y \right)}dx = {x}^{2}dy\end{array})

(\frac{d}{dx}\left( y \right) = \frac{-y\left( x+y \right)}{{{x}^{2}}})

Since both xy + y^2 and x^2 are homogeneous.

Putting, $$y = vx$$

$$\frac{dy}{dx} = v + x \frac{dv}{dx}$$

(\begin{array}{l}v\frac{dv}{dx} + x\frac{dv}{dx} = -\left( \frac{v{{x}^{2}}+{{v}^{2}}{{x}^{2}}}{{{x}^{2}}} \right)\end{array})

(\begin{array}{l}x\frac{dv}{dx} + v = -{{v}^{2}} - v\end{array})

(\int{\frac{1}{{{v}^{2}}+2v}dv = -\int{\frac{1}{v}}dv)

(\begin{array}{l}\log \left| \frac{v+1-1}{v+1+1} \right|=-2\log x+2\log c\end{array} )

(\begin{array}{l}\log \left| \frac{v}{v+2} \right| + 2\log x = 2\log c\end{array})

(\log \left| \frac{v{{x}^{2}}}{v+2} \right| = 2\log k)

(\left| k \right| = \left| \frac{v{{x}^{2}}}{v+2} \right|)

(\left| k \right| = \left| \frac{{{x}^{2}}y}{y+2x} \right|)

y = 1 and x = 1

k = 1/3

3x + 2 = y

(\begin{array}{l}\frac{2x}{3{{x}^{2}}-1}=y \ \text{which is the required solution}.\end{array} )

**Solution:**Solve (\int {{x}^{2}},\mathrm{d}x-\int \left( {{x}^{3}}+{{y}^{3}} \right),\mathrm{d}y=0)

Given:

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Solution:

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(\frac{d}{dx}\left( \frac{x^2y}{x^3+y^3} \right)=\frac{x^2y\left( 3x^2+3y^2 \right)-\left( x^3+y^3 \right)\left( 2xy \right)}{\left( x^3+y^3 \right)^2})

Putting (y = vx) and (\frac{dy}{dx} = v + x\frac{dv}{dx}),

(\begin{array}{l}v\frac{dx}{dx} + x\frac{dv}{dx} = \frac{v{{x}^{3}}}{{{x}^{3}}+{{v}^{3}}{{x}^{3}}}\end{array})

(\frac{v{x}^{3}}{{x}^{3}\left( 1+{v}^{3} \right)})

(\frac{v}{1 + v^3})

(\begin{array}{l}x\frac{dv}{dx}=\frac{v}{1+{{v}^{3}}} - v\end{array})

\(\frac{d}{dx}\left( x\left( 1+{{v}^{3}} \right) \right) = -{{v}^{4}}\)

(\left( \frac{d}{dv}\left( \frac{1+{{v}^{3}}}{{{v}^{4}}} \right) \right)dv=\frac{-dx}{x})

(\begin{array}{l} \frac{1}{3{{v}^{3}}}+\log \left| v \right|+\log \left| x \right|=c \end{array})

(\begin{array}{l} \frac{1}{3},{{x}^{3}},{{y}^{-3}}+\log \left| \frac{y}{x}\cdot x \right|=c \end{array})

$$\frac{-{{x}^{3}}}{3{{y}^{3}}}+\log \left| y \right|=c$$ is the required solution.

We solve by y = tx and proceeding accordingly.

Linear Differential Equation

(dy/dx) + py = θ

A linear differential equation: A differential equation is linear if the dependent variable (y) and its derivatives only appear to the first degree.

(\frac{dy}{dx} + P \cdot y = Q)

Here, P and Q are constants.

(\begin{array}{l}\text{We can solve this type of equation by raising both sides to the power of } {{e}^{\int{Pdx}}}\end{array})

So

(\begin{array}{l}{{e}^{\int{P, dx}}}\left( \frac{d y}{d x} + P y \right) = Q,{{e}^{\int{P, dx}}}\end{array})

(\frac{d}{dx}\left{ y,{{e}^{\int{Pdx}}} \right}=Q,{{e}^{\int{Pdx}}})

Integrating both sides,

(\begin{array}{l}\int{,Q,,{{e}^{\int{Pdx}}}dx}+c=y,{{e}^{\int{Pdx}}}\end{array} ) I said

I said, “Here.”

The expression $\left(e^{\int{Pdx}}\right)$ is called an integrating factor.

(\begin{array}{l}y[I.F] = \int{Q,(I.F),dx} + c\end{array})

Applications of Differential Equations

To Calculate Rate Measurement

To find the tangent to a curve

Illustrations:

Illustration 1: If the population of a bacteria gets tripled in 15 years, given that rate of growth is proportional to the number of bacteria, in how many years will it double?

Solution: Let the number of bacteria be x

(\therefore \frac{dx}{dt} \sim x)

(\therefore \frac{dx}{dt} = kx)

(\int{\frac{dx}{x}} = k,t + C)

(\begin{array}{l} x=e^{kt+C} \end{array})

‘(\begin{array}{l}x={{x}_{0}}{{e}^{kt}}\Rightarrow ,,x={{e}^{kt+C}}\end{array} )’

Where $$eC = x_0$$ is the population at $$t = 0$$, at $$t = 10$$ years, $$x = 2x_0$$.

(\begin{array}{l}2{{x}_{0}}={{x}_{0}},,{{e}^{k\times 10}}\end{array})

(\begin{array}{l}k=\frac{\log_{10} 2}{10}\end{array})

Now, for x = 3x⁰

t = ?

(\begin{array}{l}3{{x}_{0}}={{x}_{0}} \\ \Rightarrow t=10\frac{\log 3}{\log 2} \\ \text{where }{{e}^{\left( \frac{\log 2}{10} \right)t}} \end{array})

t = 15.9 years

Illustration 2: A biologist puts 100 bacteria in a favourable growth medium at time t = 0. After 6 hours, the number of bacteria has increased to 450. Assuming exponential growth, find the growth constant k.

Solution: $$A = P e^{kt}$$

$$450=100e^{6k}$$

(k=\frac{\ell n,4.5}{6})

1st Order Differential Equations: $\frac{dN}{dt} = kN$

Illustration 3: Five mice out of a stable population of 500 are infected with a disease in order to test a theory which states that the rate of change of the infected population is equal to α, the product of the diseased and disease-free populations.

What is the estimated incubation period for the disease?

Given:

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Solution:

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$$\ell n\left( \frac{N}{500-N} \right)=500\left( kt+c \right)$$

$$\frac{5}{500-5} = c_1 e^{500kt},\ at\ t=0,\ N=5$$

(\frac{5}{495}={{c}_{1}}{{e}^{500k\left( 0 \right)}}\Rightarrow {{c}_{1}}=\frac{1}{99})

(\therefore N = 500\left(1-\frac{1}{99},{{e}^{500kt}}\right))

$$or \begin{array}{l}N=250 \Rightarrow 1=\frac{1}{99}{{e}^{500k{{T}_{1/2}}}}\end{array} $$

$$\begin{array}{l}{T_{1/2}}=\frac{\ln\left(99\right)}{500k}=\frac{0.0091}{k}\end{array} \ \text{time units.}$$

Newton’s Cooling Law

The rate of change of temperature of a body with respect to the difference in temperature between the body and its surrounding medium is $\alpha$.

(\begin{array}{l} \frac{dT}{dt} = -k \left( T - T_m \right) \end{array})

Here’s Tm → Surrounding Temp

Cooling produces a negative change in temperature over time (dT/dt → -ve), with the temperature being greater than the melting point (T > Tm).

Illustration 5: A metal bar at 100°F at room temperature 0°F, after 20 minutes the temperature of the bar is 50°F.

(a) Time to reach 25°F

Temp after 10 min.

Solution: (\begin{array}{l}\frac{dT}{dt}+kT=-k{{T}_{m}}\Rightarrow \frac{dT}{dt}+kT=0\end{array} ) or (\begin{array}{l}{{T}_{m}}=-k\end{array} )

(or\ \begin{array}{l}T = c \cdot e^{-kt} \end{array})

T = 100 at t = 0

(\therefore c=100,{{e}^{k0}}) and

(\begin{array}{l}T=100,{{e}^{-kt}}\end{array})

At t = 20 and T = 50

(\therefore \frac{1}{2}={{e}^{-k,20}},,,,,\frac{\ln 2}{20}=k) he said

He said, “Now.”

(a) (\begin{array}{l}100=25,,{{e}^{\frac{\ell n2}{20}t}}\end{array} )

(\ell n,4=\frac{\ell {{n}^{2}}}{20}t)

or t = 40

(b) (\displaystyle T = \frac{100}{\sqrt{2}} = 70.71^\circ F)

First Order Differential Equations

Below are some of the most important and popular methods to find the solution to first-order and first-degree differential equations, along with examples:

Methods of Solving

Integrating both sides of the differential equation dy/dx = f(x) gives the general solution.

Example:

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Answer:

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(\frac{d}{dx}\left(x\right)=\frac{1}{{{x}^{2}}+1})

Given:

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Solution:

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\(\frac{dy}{dx}=\frac{x}{{{x}^{2}}+1}\)

$\int{dy} = \int{\frac{xdx}{{{x}^{2}}+1}}$

(\int{dy}=\frac{1}{2}\int{\frac{2x,,dx}{{{x}^{2}}+1}})

The solution of the given differential equation is $$y=\frac{1}{2}\log \left| {{x}^{2}}+1 \right|+c;,,,,,,,,x\in R$$

Example:

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This is an example.

(\frac{d}{dx}\left(\frac{3{{e}^{2x}}+3{{e}^{4x}}}{{{e}^{x}}+{{e}^{-x}}}\right) = \frac{6{{e}^{2x}}+12{{e}^{4x}}({{e}^{x}}-{{e}^{-x}})}{{{\left({{e}^{x}}+{{e}^{-x}}\right)}^{2}}})

Given:

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Solution:

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(\frac{d}{dx}\left( \frac{3{{e}^{2x}}+3{{r}^{4x}}}{{{e}^{x}}+\frac{1}{{{e}^{x}}}} \right)=\frac{3{{e}^{2x}}\left( 1+{{e}^{2x}} \right)}{{{e}^{x}}+\frac{1}{{{e}^{x}}}})

(\frac{dy}{dx} = 3e^{3x})

$$\int{dy}=\int{3{{e}^{3x}}dx}$$

The solution to (y=3\left( \frac{{{e}^{3x}}}{3} \right)+c) is required.

Differential Equations Reducible to Variable Separable Type

When a first order differential equation cannot be solved by integration, then we apply the substitution method in order to reduce the number of variables.

(\frac{dy}{dx} = f(px + qy))

Substitute p for x, q for y, and t for the result

For example: Solve $$\frac{dy}{dx}=\left( ax+by \right)$$

Given:

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Solution:

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(\begin{array}{l}ax + by = t\end{array})

\(\begin{array}{l}a + b \frac{dy}{dx} = \frac{dt}{dx} \end{array}\)

(\frac{dy}{dx} = \frac{1}{b}\left( \frac{dt}{dx} - a \right))

(\begin{array}{l}\frac{1}{b}\left( \frac{d t}{d x} - a \right) = t\end{array})

\(\frac{d}{dx}\left(t\right)=a+bt\)

(\int{dx}=\frac{1}{b}\int{\frac{bdt}{bt+a}})

(\begin{array}{l}x = \ln \left( bt + a \right)^{1/b}\end{array})

(\begin{array}{l}x=\ln \left( b\left( ax+by \right)+a \right)^{1/b}\end{array} )

3. Homogeneous First-Order Differential Equations:

Differential equation of the form: $$\frac{dy}{dx}=g\left( x,y \right)$$

These equations of the form (\frac{dy}{dx}=f\left( \frac{y}{x} \right)) are known as homogeneous form.

For such equations, substitute $y/x = t$

y = xt

(\frac{d}{dx}\left(y\right)=t+x\frac{d}{dx}\left(t\right))

(\begin{array}{l} \Rightarrow \frac{d}{dx}(t+x) = f(t) \end{array})

(\Rightarrow \frac{dx}{dt}=f\left( t \right)-t)

(\begin{array}{l}\int{\frac{dt}{f\left( t \right)-t}}=\int{\frac{dx}{x}}\end{array} )

Then, ’t’ can be replaced with either y or x to get a solution of the differential equation.

Answer: Find the solution to the differential equation

(\begin{array}{l} \frac{dy}{dx} = \frac{3x + y}{x - y} \end{array})

This is a great way to start a new project!

Solution: Yes, it definitely is!

(\frac{dy}{dx}=g\left( x,y \right)=\frac{3x+y}{x-y})

(\begin{array}{l}f\left( \frac{y}{x} \right) = \frac{3x + y}{x - y}\end{array})

(\frac{y}{x} = t)

(\begin{array}{l}y=xt \Rightarrow f\left(x\right)=\frac{3+x}{1-x}\end{array})

(\frac{d^2y}{dx^2} = \frac{d}{dx}\left( t + x\frac{dt}{dx} \right) = \frac{dt}{dx} + x\frac{d^2t}{dx^2} = f’(t) + xf’’(t))

(\begin{array}{l}\Rightarrow ,,,\frac{dt}{dx}=\frac{f\left( t \right)-t}{x}\end{array} )

‘(\frac{3 + 3t - t^2}{1 - t})’

(\frac{3+{{t}^{2}}}{1-t})

(\frac{dx}{dt}=\frac{1-t}{3+{{t}^{2}}})

(\int{\frac{dx}{x}} = \int{\frac{1-t}{3+t^2}dt})

$$\ln x + C_1 = \int \frac{dt}{3 + t^2} - \frac{1}{2} \int \frac{2t}{3 + t^2} , dt$$

(\frac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{t}{\sqrt{3}} \right) - \frac{1}{2}\ln \left( 3 + {{t}^{2}} \right) + {{C}_{2}})

$\begin{array}{l} \ln x + C_1 = \frac{1}{\sqrt{3}} \arctan \left( \frac{y}{\sqrt{3}x} \right) - \frac{1}{2} \ln \left( 3 + \frac{y^2}{x^2} \right) + C_2 \end{array}$

Differential Equations Practice Problems

Example 1: Solve (\frac{dy}{dx} + y = 1)

Given:

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Solution:

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$$\frac{dy}{dx} = 1 - y$$

(\left( \frac{dx}{dy} = \frac{1}{1-y} \right))

(\begin{array}{l}\int{dx}=\int{\frac{dy}{1-y}}\end{array} )

x = log |1 - y| + c is the required solution.

Example 2: Solve (\frac{dy}{dx} = \sec y)

Given:

What is the capital of France?

Solutions:

The capital of France is Paris.

(\frac{dy}{dx}=\sec y=\frac{1}{\cos y})

(\int{dx} = \int{\cos ,y,dy})

The required solution is x = sin y + c.

Example 3: Solve (\dfrac{dy}{dx} = 2xy ) when (x + 1 = 0)

Given:

What is the best way to learn a language?

Solutions:

The best way to learn a language is to immerse yourself in it. Practice speaking it as much as possible, listen to native speakers, and use language-learning tools such as apps, books, and online courses.

(\left( x+1 \right)dy = 2xy,,dx)

(\begin{array}{l}\frac{dy}{dx}=\frac{2x}{x+1}\end{array})

\( \begin{array}{l} \frac{dy}{y} = \frac{2x \, dx}{x+1} \end{array} \)

(\int{\frac{dy}{y}}=\int{\frac{2x,,dx}{\left( x+1 \right)}})

The solution to $\log ,y=2\left{ x-\log \left| x+1 \right| \right}+c$ is given.

Example 4: Solve (\int{{e}^{x}}\sqrt{1-{{y}^{2}},}dx+\frac{y}{x}dy=0 )

Given:

What are the benefits of using a CMS?

Solutions:

Benefits of using a CMS include:

• Easier website maintenance
• Enhanced user experience
• Increased security
• Improved scalability
• Increased collaboration

\(\int x{{e}^{x}}dx = \int \frac{-y}{\sqrt{1-{{y}^{2}}} } dy\)

Apply the product rule of integration

(\begin{array}{l}\int{{e}^{x}}dx - x{{e}^{x}} = \frac{1}{2}\int{\frac{dt}{\sqrt{t}}} - (1-{{y}^{2}})\end{array} )

$xe^x - e^x = \sqrt{t} + c$

$xe^x - e^x = \sqrt{1 + y^2} + c$ is the required solution.

Example 5: Solve (\frac{dy}{dx} = {{e}^{x+y}})

Given:

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Solution:

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\(\frac{dy}{dx} = e^x e^y\)

$ dy = e^x \cdot e^y \cdot dx $

(\frac{d y}{e^y} = e^x , dx)

(\int{{e}^{-y}},dy = -{e}^{-y} + C)

(\int{{e}^{x}},dx = {e}^{x} + C)

(\frac{{{e}^{x}}+C}{-1}={{e}^{-y}})

(\begin{array}{l}{e}^{x}={e}^{-y}+C\end{array})

Example 6: Solve (\frac{d}{dx}\left( {{\left( x+y \right)}^{2}} \right) ={{a}^{2}})

Given:

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Solution:

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x + y = v

Then $$\frac{dy}{dx} + 1 = \frac{dv}{dx}$$

(\begin{array}{l}\frac{dv}{dx}-1=\frac{dy}{dx} \Rightarrow \frac{dv}{dx}=\frac{dy}{dx}+1\end{array} )

$D.E. \quad {{V}^{2}}\left( \frac{dv}{dx}-1 \right)={{a}^{2} \quad }$

(\begin{array}{l}{{v}^{2}}\frac{d^2v}{dx^2}-{{v}^{2}}={{a}^{2}}\end{array})

(\begin{array}{l}{v}^{2}\frac{dv}{dx} = {a}^{2} + {v}^{2}\end{array})

(\int{\left( \frac{{{V}^{2}}}{{{a}^{2}}+{{v}^{2}}} \right),dv=x+C})

(\begin{array}{l}{\tan }^{-1}\left( \frac{v}{a} \right)=x+c,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,v-a\end{array})

(\left( x+y \right)-a\arctan \left( \frac{x+y}{a} \right)=x+C) is the required solution.

Example 7: Solve (\frac{d y}{d x} - \frac{y}{x} = 2x^2)

Given Text:

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Solution:

Hello World

(\frac{d}{dx}\left( y \right)+\left( -\frac{1}{x} \right)\left( y \right)=2{{x}^{2}})

$$P=\frac{-1}{x} \text{ and } Q=2{{x}^{2}}$$

Divide both sides by I.F.

(\begin{array}{l}I.F.={{e}^{-\int{\frac{1}{x}dx}}}={{e}^{-\log x}}\end{array})

(\begin{array}{l}={{x}^{-1}}\end{array})

(\frac{1}{x})

(\frac{d}{dx}\left(\frac{1}{x}y\right)-\frac{1}{{{x}^{2}}}y=2x)

Integrating both sides with respect to $x$, we get:

$$y\left(\frac{1}{x}\right) = \int 2x,dx + c$$

(\frac{y}{x} = x^2 + c)

(\begin{array}{l}y={{x}^{3}}+cx \quad \text{is the required solution.}\end{array} )

Important Topics in Differential Equations

![Differential-Equations - Important Topics]()

Important Questions on Differential Equations

![Differential-Equations - Important Questions]()

Differential Calculus

Top JEE Advanced Questions on Differential Equations

Frequently Asked Questions

A simple example of a differential equation is $\frac{dy}{dx} = x^2$

An example of a differential equation is: $$\frac{d^2y}{dx^2} - \frac{dy}{dx} - 3 = 0$$

A differential equation is an equation that relates a function with its derivatives. It can be written in the form of an equation involving derivatives of arbitrary order, such as a first-order differential equation, a second-order differential equation, and so on.

The order of a differential equation is the highest order derivative of the dependent variable with respect to the independent variable involved in the given differential equation.

A degree of a differential equation is the highest power of the derivative present in the equation.

The degree of the differential equation is the power of the highest order derivative present in the equation.

1. Modeling the motion of objects in physics
2. Modeling population growth in biology

Newton’s law of fall of an object and Newton’s law of cooling are two examples of applications of differential equations.