Dr. Brajesh Kumar Jha

Differential Equations: Detailed Step-by-Step Solutions In this article, we solve three important differential equations step by step using standard methods such as complementary function and particular integral , integrating factor , and separation of variables . 1) Solve \(y''+4y=\sin(3x)\) This is a linear differential equation with constant coefficients: \[ y''+4y=\sin(3x) \] The general solution is: \[ y=y_c+y_p \] where \(y_c\) is the complementary function and \(y_p\) is the particular integral. Step 1: Complementary Function The auxiliary equation is: \[ m^2+4=0 \] \[ m^2=-4 \] \[ m=\pm 2i \] Therefore, the complementary function is: \[ y_c=C_1\cos 2x+C_2\sin 2x \] Step 2: Particular Integral \[ (D^2+4)y=\sin 3x \] \[ y_p=\frac{1}{D^2+4}\sin 3x \] Using: \[ f(D)\sin ax=f(-a^2)\sin ax \] \[ y_p=\frac{1}{-9+4}\sin 3x \] \[ y_p=-\frac15\sin 3x \] Step 3: General Solution \[ y=C_1\cos 2x+C_2\sin 2x-\frac15\sin 3x \] Fi...

First Order Homogeneous Differential Equations (Solved Examples) A differential equation of the form \[ \frac{dy}{dx}=f\left(\frac{y}{x}\right) \] is called a homogeneous differential equation . We use the substitution \[ y=vx \quad \text{or} \quad v=\frac{y}{x} \] so that \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Example 1: Solve \( \frac{dy}{dx}=\frac{x+y}{x} \) View Solution \[ \frac{dy}{dx}=1+\frac{y}{x} \] Let \[ v=\frac{y}{x}, \quad y=vx \] \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Substitute \[ v+x\frac{dv}{dx}=1+v \] \[ x\frac{dv}{dx}=1 \] \[ \frac{dv}{dx}=\frac{1}{x} \] Integrate \[ v=\ln|x|+C \] Since \(v=\frac{y}{x}\) \[ \frac{y}{x}=\ln|x|+C \] \[ y=x(\ln|x|+C) \] Example 2: Solve \( \frac{dy}{dx}=\frac{x+y}{x-y} \) View Solution Let \[ v=\frac{y}{x}, \quad y=vx \] \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Substitute \[ v+x\frac{dv}{dx}=\frac{1+v}{1-v} \] \[ x\frac{dv}{dx}=\frac{1+v}{1-v}-v \] \[ x\frac{dv}{dx}=\frac{1+v^2}{...

First Order Differential Equations using Separation of Variables Below are solved examples of first order differential equations using the separation of variables method . Example 1: Solve \( \frac{dy}{dx}=3x^2 \) View Solution Separate variables \[ dy = 3x^2 dx \] Integrate both sides \[ \int dy = \int 3x^2 dx \] \[ y = x^3 + C \] Example 2: Solve \( \frac{dy}{dx}=xy \) View Solution Separate variables \[ \frac{dy}{y}=x\,dx \] Integrate \[ \int \frac{1}{y}dy = \int x\,dx \] \[ \ln |y| = \frac{x^2}{2}+C \] \[ y = Ce^{x^2/2} \] Example 3: Solve \( \frac{dy}{dx}=\frac{x}{y} \) View Solution Separate variables \[ y\,dy = x\,dx \] Integrate \[ \int y\,dy = \int x\,dx \] \[ \frac{y^2}{2} = \frac{x^2}{2} + C \] \[ y^2 = x^2 + C \] Example 4: Solve \( \frac{dy}{dx}=y^2 \) View Solution Separate variables \[ \frac{dy}{y^2}=dx \] Integrate \[ \int y^{-2}dy = \int dx \] \[ -\frac{1}{y}=x+C \] \[ y=\frac{1}{C-x} ...

First Order Differential Equations using Separation of Variables Advanced Application Problems (Separation of Variables) The following examples illustrate applications of first order differential equations in population growth, cooling law, chemical reactions, and physics. Example 11 (Population Growth): If population grows according to \( \frac{dP}{dt}=kP \), find \(P(t)\). View Solution Separate variables \[ \frac{dP}{P}=k\,dt \] Integrate \[ \int \frac{dP}{P}=\int k\,dt \] \[ \ln P = kt + C \] \[ P = Ce^{kt} \] Example 12 (Newton's Law of Cooling): \[ \frac{dT}{dt}=-k(T-T_s) \] Find \(T(t)\). View Solution Separate variables \[ \frac{dT}{T-T_s}=-k\,dt \] Integrate \[ \ln |T-T_s|=-kt+C \] \[ T-T_s=Ce^{-kt} \] \[ T=T_s+Ce^{-kt} \] Example 13 (Radioactive Decay): Solve \( \frac{dN}{dt}=-kN \) View Solution \[ \frac{dN}{N}=-k\,dt \] Integrate \[ \ln N=-kt+C \] \[ N=Ce^{-kt} \] Example 14 (Chemical Reaction): So...

Cholesky Decomposition Method – Solved Problems The Cholesky Decomposition Method is used to solve systems of linear equations when the coefficient matrix is symmetric and positive definite . The matrix is decomposed as A = LL T where L is a lower triangular matrix. The solution is obtained in two steps: Forward substitution: LY = B Backward substitution: L T X = Y Problem 1 Solve using Cholesky decomposition 4x + 2y = 6 2x + 3y = 7 Show Solution Matrix form A = [4 2 2 3] L = [2 0 1 √2] Forward substitution: 2y₁ = 6 y₁ = 3 3 + √2 y₂ = 7 y₂ = 2√2 Backward substitution: √2 y = 2√2 y = 2 2x + 2 = 3 x = 1/2 Solution: x = 1/2 y = 2 Problem 2 Solve 4x + 2y + 2z = 8 2x + 5y + z = 3 2x + y + 3z = 5 Show Solution Matrix A = [4 2 2 2 5 1 2 1 3] Cholesky factor L = [2 0 0 1 2 0 1 0 √2] Forward substitution gives Y = [4 -1/2 1/√2] Backwar...

Total Derivatives – Theory and Solved Problems Definition If \(z=f(x,y)\) where \(x\) and \(y\) depend on parameter \(t\), then the total derivative is \[ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \] This formula is known as the **Chain Rule for multivariable functions**. Visual Understanding of the Chain Rule In problems involving total derivatives , variables often depend on other variables. The Chain Rule shows how a change in one variable affects another through intermediate variables. The following diagrams illustrate this dependency structure. Diagram 1: Basic Chain Rule Structure z x y t This diagram represents: \[ z = f(x,y), \quad x = x(t), \quad y = y(t) \] \[ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \] Diagram 2: Two-Level Dependency z x y u...

Double Integration in Polar Coordinates Animated Step-by-Step Solutions Example 1: Area of Circle $x^2+y^2 \le 4$ Evaluate: $$\iint_R 1 \, dA$$ ▶ Show Detailed Solution Step 1: Convert to Polar Coordinates $x = r\cos\theta$, $y = r\sin\theta$ $x^2+y^2 = r^2$ $dA = r\,dr\,d\theta$ Step 2: Limits $0 \le r \le 2$, $0 \le \theta \le 2\pi$ Step 3: Integral $$ \int_0^{2\pi}\int_0^2 r\,dr\,d\theta $$ $$ \int_0^2 r\,dr = \frac{r^2}{2}\Big|_0^2 = 2 $$ $$ \int_0^{2\pi} 2\,d\theta = 4\pi $$ Final Answer: $4\pi$ Example 2: Quarter Circle Integral Evaluate: $$\iint_R (x^2+y^2)\, dA$$ ▶ Show Detailed Solution Step 1: $x^2+y^2 = r^2$ Limits: $0 \le r \le 3$ $0 \le \theta \le \frac{\pi}{2}$ $$ \int_0^{\pi/2}\int_0^3 r^3\,dr\,d\theta $$ $$ \int_0^3 r^3 dr = \frac{81}{4} $$ $$ \frac{81}{4} \cdot \frac{\pi}{2} = \frac{81\pi}{8} $$ Final Answer: $\frac{81\pi}{8}$ Example 3: Annular Region Evaluate: $$\iint_R 1\,dA$$ ▶ ...