The family of curves is: \(y=Ae^{2x}+Be^{-x}+Csin⁡x\) Here, the arbitrary constants are: \(A\) \(B\) \(C\) Thus, there are 3 independent arbitrary constants. To form the differential equation, the relation is differentiated repeatedly until all three constants are eliminated. Consequently: The resulting differential equation is of third order. The general solution of a third-order differential equation always contains three arbitrary constants. Hence, Option C is correct.

(1) \(y=asin⁡x+bcos⁡x\)→ 2 constants → second-order family structure → (c) (2) \(y=mx+c\)→ linear first-order family → (d) (3) \(f(x,y,c_{1},c_{2})=0\)→ two constants → second-order general solution → (a) (4) \(f(x,y,c)=0\)→ one constant → first-order general solution → (b) Thus correct matching is (1)-(c), (2)-(d), (3)-(a), (4)-(b)

A is correct: constants are fixed using conditions B is correct: final solution has no arbitrary constants C is incorrect in concept but often included as exam trap; however in strict NCERT logic it is false because degree does not determine constants, so option C should NOT be included So correct set is A, B only → Option B in given pattern, but since provided key forces grouping, correct interpretation is A, B only.

A particular solution is obtained after substituting initial or boundary conditions, which uniquely determine all arbitrary constants. Therefore, no arbitrary constant remains in the final expression.

The given family of curves is: \(x^{2}+y^{2}=a^{2}\) where \(a\) is the arbitrary constant. Differentiate both sides with respect to \(x\): \(\frac{d}{dx}(x^{2}+y^{2})=\frac{d}{dx}(a^{2})2x+2y\frac{dy}{dx}=0\) Dividing throughout by 2, \(x+y\frac{dy}{dx}=0\) This differential equation represents the entire family of circles centered at the origin. Hence, Option A is correct.

To solve the differential equation: \(\frac{dy}{dx}=y\) Separate the variables: \(\frac{dy}{y}=dx\) Integrate both sides: \(\int \frac{1}{y} dy=\int dxln⁡∣y∣=x+C\) Exponentiating, \(y=Ae^{x},\) where \(A=e^{C}\)is an arbitrary constant. Since the problem specifies \(y>0\), we have \(A>0\). Thus, every integral curve is an exponentially growing curve, increasing continuously as \(x\) increases. Hence, Option C is correct.

Given, \(y=5e^{-x}\) Differentiate both sides with respect to \(x\): \(\frac{dy}{dx}=5⋅\frac{d}{dx}(e^{-x})\) Using the chain rule, \(\frac{d}{dx}(e^{-x})=-e^{-x}\) Therefore, \(\frac{dy}{dx}=5(-e^{-x})=-5e^{-x}\) Verification: \(\frac{dy}{dx}+y=-5e^{-x}+5e^{-x}=0\) Hence, \(y=5e^{-x}\)satisfies the differential equation. Therefore, Option B is correct.

Given, \(y=x^{2}log⁡x-\frac{x^{2}}{2}+c\) To verify the solution, differentiate both sides. The first term, \(x^{2}log⁡x,\) is a product of two functions: \(u=x^{2}\) \(v=log⁡x\) Apply the Product Rule: \(\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\) Thus, \(\frac{d}{dx}(x^{2}log⁡x)=x^{2}\left(\frac{1}{x}\right)+log⁡x(2x)=x+2xlog⁡x\) Differentiate the remaining terms: \(\frac{d}{dx}\left(,\ \frac{x^{2}}{2}\right)=-x\frac{d}{dx}(c)=0\) Therefore, \(\frac{dy}{dx}=(x+2xlog⁡x)-x=2xlog⁡x\) Since the required derivative in the differential equation is \(\frac{dy}{dx}=2xlog⁡x+x,\) the given expression for \(y\) does not satisfy that equation. (It would satisfy \(\frac{dy}{dx}=2xlog⁡x\).) Nevertheless, the primary differentiation rule used is the Product Rule, making Option C the correct answer.

Differentiating: \(2x+2y\frac{dy}{dx}=0\) Dividing by 2: \(x+y\frac{dy}{dx}=0\) This is always valid → probability = 1.

The given implicit function is: \(\left(x-2)^{2}+(y-3)^{2}=5^{2}\right.\) Differentiate both sides with respect to \(x\): \(\frac{d}{dx}\left[(x-2)^{2}\right]+\frac{d}{dx}\left[(y-3)^{2}\right]=\frac{d}{dx}(25)\) Now, \(\frac{d}{dx}\left[(x-2)^{2}\right]=2(x-2)\) Since \(y\) depends on \(x\), \(\frac{d}{dx}\left[(y-3)^{2}\right]=2(y-3)\frac{dy}{dx}\) Also, \(\frac{d}{dx}(25)=0\) Therefore, \(2(x-2)+2(y-3)\frac{dy}{dx}=0\) Hence, Option A is correct.

Given, \(\frac{dy}{dx}=\frac{x+1}{2-x}\) To integrate, \(\int \frac{x+1}{2-x} dx,\) the integrand is first simplified using algebraic manipulation: \(\frac{x+1}{2-x}=-1+\frac{3}{2-x}.\) Now the integral becomes \(\int \left(-1+\frac{3}{2-x}\right)dx,\) which is easy to evaluate: \(=-x-3ln⁡∣2-x∣+C.\) Thus, the most appropriate technique is polynomial division or algebraic manipulation, making Option C correct.

(1) y = a sin x + b cos x has two constants → second order general solution. (2) y = mx + c has one constant → first order family. (3) and (4) represent implicit general solution forms matching respective families.

Given, \(\frac{dy}{dx}=e^{x+y}+x^{2}e^{y}\) 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, with all terms involving \(y\) on the left and all terms involving \(x\) on the right. Hence, Option A is correct.

Given, \(ylog⁡y dx-x dy=0\) Rearrange the equation: \(ylog⁡y dx=x dy\) Separate the variables: \(\frac{dx}{x}=\frac{dy}{ylog⁡y}\) Now integrate both sides: \(\int \frac{dx}{x}=\int \frac{dy}{ylog⁡y}\) Using the standard integration formulas, \(\int \frac{dx}{x}=log⁡∣x∣+C_{1}\) and \(\int \frac{dy}{ylog⁡y}=log⁡∣log⁡y∣+C_{2}\) Combining the constants, \(log⁡∣x∣=log⁡∣log⁡y∣+log⁡c\) Hence, Option B is correct.

Given, \(xy\frac{dy}{dx}=(x+2)(y+2)\) Divide both sides by xy: \(\frac{dy}{dx}=\frac{\left(x+2)(y+2\right)}{xy}\) Rewrite as \(\frac{dy}{dx}=\frac{x+2}{x}\times \frac{y+2}{y}\) Now separate the variables: \(\frac{y}{y+2} dy=\frac{x+2}{x} dx\) Thus, all y-terms are on the left and all x-terms are on the right. Hence, Option A is correct.

After integrating, \(\int \frac{dx}{x}=\int \frac{dy}{ylog⁡y}\) we obtain \(log⁡∣x∣+C_{1}=log⁡∣log⁡y∣+C_{2}.\) Combining the constants, \(log⁡∣x∣=log⁡∣log⁡y∣+C,\) where C = C₂ − C₁ is an arbitrary constant. Since C is arbitrary, we may write \(C=log⁡c,\) where c > 0. Substituting, \(log⁡∣x∣=log⁡∣log⁡y∣+log⁡c.\) Using the logarithmic identity, \(log⁡a+log⁡b=log⁡(ab),\) this can also be written as \(log⁡∣x∣=log⁡(c ∣log⁡y∣).\) Thus, expressing the constant as log c is simply a convenient algebraic representation of an arbitrary constant. Hence, Option B is correct.

Given, \(\frac{dy}{dx}=\frac{1}{x}-\frac{1}{x^{2}}\) Integrate both sides: \(\int dy=\int \left(\frac{1}{x},\ \frac{1}{x^{2}}\right)dx\) Apply the standard integrals: \(\int \frac{1}{x} dx=log⁡x\) and \(\int -\frac{1}{x^{2}} dx=-\int x^{-2}dx=-\left(\frac{x^{-1}}{-1}\right)=\frac{1}{x}\) Therefore, \(y=log⁡x+\frac{1}{x}+c\) Hence, Option B is correct.

Given, \(e^{x}\sqrt{1-y^{2}} dx+\frac{y}{x} dy=0\) Move the second term to the right side: \(e^{x}\sqrt{1-y^{2}} dx=-\frac{y}{x} dy\) Multiply both sides by \(\frac{x}{\sqrt{1-y^{2}}}\) to separate the variables: \(xe^{x} dx=-\frac{y}{\sqrt{1-y^{2}}} dy\) Now, all x-terms are on the left and all y-terms are on the right. Hence, Option A is correct.

The general solution is \(y=ke^{x}-1.\) Since the curve passes through the origin, \(x=0,y=0.\) Substitute these values: \(0=ke^{0}-1\) Since \(e^{0}=1,\) we get \(0=k-1.\) Therefore, \(k=1.\) Hence, Option B is correct.

From the passage, the general solution is \(y=ke^{x}-1.\) Since the curve passes through the origin, \((x,y)=(0,0).\) Substitute these values into the equation: \(0=ke^{0}-1.\) Since \(e^{0}=1,\) we get \(0=k-1,\) so \(k=1.\) Substituting \(k=1\) into the general solution gives \(y=e^{x}-1.\) Hence, Option A is correct.