The given general solution is \(y=c_{1}e^{x}+c_{2}e^{-x}+c_{3}e^{x}.\) Notice that both c₁eˣ and c₃eˣ contain the same function eˣ. Combine them as \(c_{1}e^{x}+c_{3}e^{x}=(c_{1}+c_{3})e^{x}.\) Let \(C=c_{1}+c_{3}.\) Then the solution becomes \(y=Ce^{x}+c_{2}e^{-x}.\) Now there are only two independent arbitrary constants: C c₂ Since the order of a differential equation equals the number of independent arbitrary constants in its general solution, the resulting differential equation is of order 2. Hence, Option B is correct.

(1) a sin(x+b) → 2 parameters → order c mapping (2) circle → single differentiation → order a (3) exponential form → order b reduction (4) 4 parameters → highest order (d)

Given, \((x+1)\frac{dy}{dx}=2xy.\) Step 1: Separate the variables \(\frac{dy}{y}=\frac{2x}{x+1} dx.\) Hence, Statement A is correct. Step 2: Simplify the integrand Rewrite \(\frac{2x}{x+1}=2-\frac{2}{x+1}.\) This makes the integration straightforward. Hence, Statement C is correct. Step 3: Integrate both sides \(\int \frac{dy}{y}=\int \left(2,\ \frac{2}{x+1}\right)dx.\) Therefore, \(log⁡∣y∣=2x-2log⁡∣x+1∣+c.\) Hence, Statement B is correct. Step 4: Apply the initial condition Using \(x=2,y=3,log⁡3=4-2log⁡3+c.\) Thus, \(c=log⁡3-4+2log⁡3.\) Hence, Statement D is also correct. Therefore, all four statements are correct. Answer: Option D

Given, \(\frac{dy}{dx}=x(2log⁡x+1).\) Integrate both sides: \(y=\int x(2log⁡x+1) dx.\) Expanding, \(y=2\int xlog⁡x dx+\int x dx.\) Evaluate \(\int xlog⁡x dx\) using integration by parts. Let \(u=log⁡x,dv=x dx.\) Then, \(du=\frac{1}{x} dx,v=\frac{x^{2}}{2}.\) Applying integration by parts, \(\int xlog⁡x dx=\frac{x^{2}}{2}log⁡x-\frac{x^{2}}{4}.\) Therefore, \(y=x^{2}log⁡x+c.\) Using the initial condition \(y(2)=0,0=4log⁡2+c.\) Hence, \(c=-4log⁡2.\) Therefore, Statements A, B, and C are correct. Statement D is incorrect because \(\frac{d}{dx}(log⁡x)=\frac{1}{x},\) but \(\int log⁡x dx\neq \frac{1}{x}.\) Hence, Option D is the incorrect statement.

Given the family of curves, \(x^{2}=4ay.\) Differentiate both sides with respect to \(x\): \(2x=4a\frac{dy}{dx}.\) From the original equation, \(4a=\frac{x^{2}}{y}.\) Substitute this into the differentiated equation: \(2x=\frac{x^{2}}{y}\frac{dy}{dx}.\) Multiply both sides by \(y\): \(2xy=x^{2}\frac{dy}{dx}.\) Divide both sides by \(x\left(x,\ 0\right)\): \(2y=x\frac{dy}{dx}.\) Therefore, \(x\frac{dy}{dx}=2y.\) Hence, Option B is correct.

The given family of circles is \(x^{2}+(y-a)^{2}=a^{2}.\) Here, \(a\) is the only arbitrary constant representing the y-coordinate of the centre. To form the differential equation: 1. Differentiate the given equation with respect to \(x\). 2. Use the original equation and the differentiated equation together. 3. Eliminate the arbitrary constant \(a\). The objective is not to remove \(a\) because it is fixed, but because differential equations are formed by eliminating arbitrary constants from the given family of curves. Hence, Option B is correct.

Given, \(y=2e^{2x}-3e^{-x}.\) Step 1: Find the first derivative Differentiate term by term: \(\frac{dy}{dx}=2\left(2e^{2x}\right)-3\left(,\ e^{-x}\right)\frac{dy}{dx}=4e^{2x}+3e^{-x}.\) Step 2: Find the second derivative Differentiate again: \(\frac{d^{2}y}{dx^{2}}=4\left(2e^{2x}\right)+3\left(,\ e^{-x}\right)\frac{d^{2}y}{dx^{2}}=8e^{2x}-3e^{-x}.\) Hence, Option B is correct

Given, \(y=a+log⁡x.\) Step 1: Find the first derivative Since a is a constant, \(\frac{d}{dx}(a)=0.\) Also, \(\frac{d}{dx}(log⁡x)=\frac{1}{x}.\) Therefore, \(\frac{dy}{dx}=\frac{1}{x}.\) Step 2: Find the second derivative Differentiate again: \(\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left(\frac{1}{x}\right)=\frac{d}{dx}(x^{-1})=-x^{-2}.\) Hence, \(\frac{d^{2}y}{dx^{2}}=-\frac{1}{x^{2}}.\) Therefore, Option B is correct.

The family of circles with centre on the x-axis and passing through the origin is represented by \(\left(x-a)^{2}+y^{2}=a^{2}.\right.\) To obtain or verify the corresponding differential equation: 1. Differentiate the equation implicitly with respect to \(x\). 2. Eliminate the arbitrary constant \(a\). 3. Simplify the result to obtain \((x^{2}-y^{2})\frac{dy}{dx}-2xy=0.\) Since every verification of this family requires implicit differentiation, the event is certain. Therefore, \(P(Implicit differentiation is required)=1.\) Hence, Option D is correct.

Given, \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1.\) Step 1: Differentiate implicitly \(\frac{2x}{a^{2}}+\frac{2y}{b^{2}}\frac{dy}{dx}=0.\) Dividing by 2, \(\frac{x}{a^{2}}+\frac{y\frac{dy}{dx}}{b^{2}}=0.\) Step 2: Rearrange Move one term to the other side: \(\frac{x}{a^{2}}=-\frac{y\frac{dy}{dx}}{b^{2}}.\) Hence, \(\frac{yy_{1}}{x}=-\frac{b^{2}}{a^{2}},\) where \(y_{1}=\frac{dy}{dx}.\) The right-hand side now consists only of constants. Step 3: Differentiate again Differentiate \(\frac{yy_{1}}{x}=-\frac{b^{2}}{a^{2}}\) with respect to \(x\). Since the right-hand side is constant, its derivative is zero. This eliminates the arbitrary constants completely and leads to \(x\left(y\frac{d^{2}y}{dx^{2}},\ {\left(\frac{dy}{dx}\right)}^{2}\right)=y\frac{dy}{dx}.\) Therefore, Option B is correct.

Given, \(\frac{dy}{dx}=e^{x}.e^{y}+x^{2}e^{y}.\) Factor out \(e^{y}\): \(\frac{dy}{dx}=e^{y}(e^{x}+x^{2}).\) Now divide both sides by \(e^{y}\): \(e^{-y}\frac{dy}{dx}=e^{x}+x^{2}.\) Multiply both sides by \(dx\): \(e^{-y} dy=(e^{x}+x^{2}) dx.\) This successfully separates the variables, making the equation ready for integration. Hence, Option B is correct.

Given, \(\frac{y}{y+2} dy=\frac{x+2}{x} dx.\) Step 1: Simplify both sides Write \(\frac{y}{y+2}=1-\frac{2}{y+2}\) and \(\frac{x+2}{x}=1+\frac{2}{x}.\) Step 2: Integrate both sides \(\int \left(1,\ \frac{2}{y+2}\right)dy=\int \left(1,\ \frac{2}{x}\right)dx.\) Integrating, \(y-2log⁡∣y+2∣=x+2log⁡∣x∣+c.\) Hence, Option A is correct.

Given, \(x(e^{2y}-1) dy+(x^{2}-1)e^{y} dx=0.\) Move the second term to the right-hand side: \(x(e^{2y}-1) dy=-(x^{2}-1)e^{y} dx.\) Since \(-(x^{2}-1)=1-x^{2},\) we have \(x(e^{2y}-1) dy=(1-x^{2})e^{y} dx.\) Now divide both sides by \(xe^{y}.\) This gives \(\frac{e^{2y}-1}{e^{y}} dy=\frac{1-x^{2}}{x} dx.\) Thus, the variables are completely separated. Hence, Option A is correct.

Given, \((e^{y}-e^{-y}) dy=\left(\frac{1}{x},\ x\right) dx.\) Integrate both sides: \(\int (e^{y}-e^{-y}) dy=\int \left(\frac{1}{x},\ x\right) dx.\) Left-hand side Using standard integrals, \(\int e^{y} dy=e^{y}\) and \(\int (-e^{-y}) dy=e^{-y}.\) Therefore, \(\int (e^{y}-e^{-y}) dy=e^{y}+e^{-y}.\) Right-hand side Using standard integrals, \(\int \frac{1}{x} dx=log⁡∣x∣\) and \(\int (-x) dx=-\frac{x^{2}}{2}.\) Hence, \(\int \left(\frac{1}{x},\ x\right) dx=log⁡∣x∣-\frac{x^{2}}{2}+c.\) Therefore, \(e^{y}+e^{-y}=log⁡∣x∣-\frac{x^{2}}{2}+c.\) Hence, Option A is correct.

Given, \(log⁡\left(\frac{dy}{dx}\right)=3x+4y.\) Using the logarithmic identity \(log⁡A=B⟺A=e^{B},\) we obtain \(\frac{dy}{dx}=e^{3x+4y}.\) Further, \(e^{3x+4y}=e^{3x}e^{4y},\) which can then be separated as \(e^{-4y} dy=e^{3x} dx.\) Thus, the initial algebraic manipulation is converting the logarithmic equation into its exponential form. Hence, Option B is correct.

Given, \(\frac{e^{-4y}}{-4}=\frac{e^{3x}}{3}+c.\) Multiply both sides by −12: \(3e^{-4y}=-4e^{3x}-12c.\) Since \(c\) is an arbitrary constant, \(-12c\) is also an arbitrary constant. Therefore, instead of writing −12c, we simply write \(C\) (or \(C_{1}\)). Thus, the equation becomes \(3e^{-4y}=-4e^{3x}+C.\) This is standard practice in solving differential equations. Hence, Option B is correct.

Given, \(log⁡\left(\frac{dy}{dx}\right)=3x+4y.\) Step 1: Convert to exponential form Using \(log⁡A=B⟹A=e^{B},\) we obtain \(\frac{dy}{dx}=e^{3x+4y}=e^{3x}e^{4y}.\) Step 2: Separate the variables \(e^{-4y} dy=e^{3x} dx.\) Step 3: Integrate both sides \(\int e^{-4y} dy=\int e^{3x} dx.\) This gives \(-\frac{e^{-4y}}{4}=\frac{e^{3x}}{3}+c.\) Step 4: Eliminate the fractions Multiply both sides by −12: \(3e^{-4y}=-4e^{3x}+C.\) Rearranging, \(3e^{-4y}+4e^{3x}=C.\) Hence, Option A is correct.

Given, \(\frac{dy}{dx}=\frac{x+1}{2-y}.\) Step 1: Separate the variables \((2-y) dy=(x+1) dx.\) Step 2: Integrate both sides \(\int (2-y) dy=\int (x+1) dx.\) Evaluating the integrals, \(2y-\frac{y^{2}}{2}=\frac{x^{2}}{2}+x+c.\) Step 3: Remove the fraction Multiply both sides by 2: \(4y-y^{2}=x^{2}+2x+C.\) Rearranging, \(y^{2}-4y+x^{2}+2x=C.\) Hence, Option A is correct.

The general solution is \(y-2log⁡∣y+2∣=x+2log⁡∣x∣+c.\) Substitute \(x=1,y=-1.\) Then, \(-1-2log⁡1=1+2log⁡1+c.\) Since \(log⁡1=0,\) we get \(-1=1+c.\) Therefore, \(c=-2.\) Hence, Option B is correct.

The general solution is \(y-2log⁡∣y+2∣=x+2log⁡∣x∣+c.\) Substituting \(c=-2,\) gives \(y-2log⁡∣y+2∣=x+2log⁡∣x∣-2.\) Rearranging, \(y-x+2=2log⁡∣x∣+2log⁡∣y+2∣.\) Using the logarithmic identity \(log⁡a+log⁡b=log⁡(ab),\) we obtain \(2log⁡∣x(y+2)∣.\) Hence, \(y-x+2=2log⁡∣x(y+2)∣.\) Therefore, Option A is correct.