Q.The differential equation of the family of curves $x^2+y^2-2ay=0$, where $a$ is arbitrary constant, is:
(A) $(x^2-y^2)\frac{dy}{dx}=2xy$
(B) $2(x^2+y^2)\frac{dy}{dx}=xy$
(C) $2(x^2-y^2)\frac{dy}{dx}=xy$
(D) $(x^2+y^2)\frac{dy}{dx}=2xy$

Showing the 12 most recent of 20 on this concept.

1. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.The differential equation of the family of all circles of radius ‘a’ is (A) $y_2^2 + 1 = y_1^2 + a^2$ (B) $1 + y_1^2 = y_2^2 + a^2$ (C) $y_1 y_2 + (1 + y_1^2) = a$ (D) $(1 + y_1^2)^3 = a^2 y_2^2$

   ›Reveal solutionSolution

   The key idea is to start with the general equation of a circle of fixed radius $a$, eliminate the two arbitrary constants (centre coordinates) by differentiating twice, and then simplify to obtain the differential equation. The correct result is $(1 + y_1^2)^3 = a^2 y_2^2$, which matches option (D).

   The problem asks for the differential equation that represents all circles of a given radius $a$, regardless of where their centre lies. That means the family has two free parameters — the $x$ and $y$ coordinates of the centre. To eliminate them, we need to differentiate the circle’s equation twice, because each differentiation removes one constant. The final relation between $y$, $y_1$ (first derivative), and $y_2$ (second derivative) must be free of the centre coordinates.

   Let’s work through it.

   1. Write the general equation of a circle of radius $a$. Let the centre be at $(h, k)$. Then

   $(x - h)^2 + (y - k)^2 = a^2.$

   Here $h$ and $k$ are the arbitrary constants we need to eliminate.

   2. Differentiate once with respect to $x$. Using the chain rule:

   $2(x - h) + 2(y - k) y_1 = 0,$

   where $y_1 = \frac{dy}{dx}$. Divide through by 2:

   $(x - h) + (y - k) y_1 = 0. \qquad(1)$

   3. Differentiate a second time. Differentiate (1) with respect to $x$:

   $1 + (y - k) y_2 + y_1 \cdot y_1 = 0,$

   because the derivative of $(y - k) y_1$ is $(y - k) y_2 + y_1^2$. So

   $1 + (y - k) y_2 + y_1^2 = 0. \qquad(2)$

   4. Eliminate $(y - k)$ from (1) and (2). From (1), we have $(x - h) = - (y - k) y_1$. But we don’t need $x - h$ directly — we need to get rid of $(y - k)$. From (2):

   $(y - k) y_2 = - (1 + y_1^2).$

   So

   $y - k = -\frac{1 + y_1^2}{y_2}, \quad \text{provided } y_2 \neq 0.$

   5. Now eliminate $x - h$ using (1). From (1): $x - h = - (y - k) y_1$. Substitute the expression for $y - k$:

   $x - h = - \left( -\frac{1 + y_1^2}{y_2} \right) y_1 = \frac{y_1 (1 + y_1^2)}{y_2}.$

   6. Plug both $x - h$ and $y - k$ back into the original circle equation. The original equation is $(x - h)^2 + (y - k)^2 = a^2$. Substituting:

   $\left( \frac{y_1 (1 + y_1^2)}{y_2} \right)^2 + \left( -\frac{1 + y_1^2}{y_2} \right)^2 = a^2.$

   Factor $(1 + y_1^2)^2 / y_2^2$ out of both terms:

   $\frac{(1 + y_1^2)^2}{y_2^2} \left( y_1^2 + 1 \right) = a^2.$ …

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2. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The differential equation of the family of all circles of radius ‘a’ is (A) $y_1 y_2 + (1 + y_1^2) = a$ (B) $(1 + y_1^2)^3 = a^2 y_2^2$ (C) $1 + y_1^2 = y_2^2 + a^2$ (D) $y_2^2 + 1 = y_1^2 + a^2$

   ›Reveal solutionSolution

   The family of all circles of fixed radius $a$ has a differential equation that eliminates the two arbitrary parameters (center coordinates). The correct equation is $(1 + y_1^2)^3 = a^2 y_2^2$, which is option (B).

   We start with the general equation of a circle of radius $a$:

   $(x - h)^2 + (y - k)^2 = a^2$

   Here $h$ and $k$ are the coordinates of the center — two arbitrary constants. To get a differential equation that describes all such circles (no matter where they are placed), we must eliminate $h$ and $k$ by differentiating.

   Why this approach works:

   Each differentiation reduces the number of arbitrary constants. Two constants require two derivatives. The resulting relation between $y, y_1, y_2$ (where $y_1 = dy/dx$, $y_2 = d^2y/dx^2$) will be free of $h$ and $k$ and will involve only $a$.

   ---

   1. First derivative Differentiate the circle equation implicitly with respect to $x$:

   $2(x - h) + 2(y - k) y_1 = 0$

   Divide by 2:

   $(x - h) + (y - k) y_1 = 0 \quad \text{(1)}$

   This gives a linear relation between $x - h$ and $y - k$.

   2. Second derivative Differentiate (1) again with respect to $x$:

   $1 + (y - k) y_2 + y_1^2 = 0$

   (Remember: derivative of $(y - k) y_1$ is $y_1 \cdot y_1 + (y - k) y_2 = y_1^2 + (y - k) y_2$.)

   So:

   $1 + y_1^2 + (y - k) y_2 = 0 \quad \text{(2)}$

   3. Eliminate $y - k$ From (2):

   $y - k = -\frac{1 + y_1^2}{y_2}$

   Substitute into (1):

   $x - h + \left(-\frac{1 + y_1^2}{y_2}\right) y_1 = 0$

   So:

   $x - h = \frac{y_1(1 + y_1^2)}{y_2}$

   4. Use the original circle equation Plug $x - h$ and $y - k$ into $(x - h)^2 + (y - k)^2 = a^2$: $$$$

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3. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.The general solution of the differential equation $(3x^2-2xy)dy+(y^2-2xy)dx=0$ is (A) $x^2-xy=cy^2$ (B) $y^2-xy=cx^3$ (C) $xy-x^2=cy^3$ (D) $xy-y^2=cx^3$

   ›Reveal solutionSolution

   This is a homogeneous differential equation. Substituting $y = vx$ reduces it to a separable form, leading to the solution $xy - x^2 = c\,y^3$, which corresponds to option (C).

   We start by recognizing the structure: the equation

   $(3x^2 - 2xy)\,dy + (y^2 - 2xy)\,dx = 0$

   has every term of total degree 2 (e.g., $3x^2$, $-2xy$, $y^2$ are all degree 2). That is the hallmark of a homogeneous differential equation — one where $M(x,y)$ and $N(x,y)$ are homogeneous functions of the same degree. For such equations, the substitution $y = vx$ turns it into a separable equation in $v$ and $x$.

   Let’s work through it step by step.

   1. Rewrite in standard form We have $M(x,y)\,dx + N(x,y)\,dy = 0$ with

   $M(x,y) = y^2 - 2xy, \qquad N(x,y) = 3x^2 - 2xy,$

   both homogeneous of degree 2.

   2. Substitute $y = vx$ Then $dy = v\,dx + x\,dv$, and

   $M = (vx)^2 - 2x(vx) = x^2(v^2 - 2v), \qquad N = 3x^2 - 2x(vx) = x^2(3 - 2v).$

   The equation becomes

   $x^2(v^2 - 2v)\,dx + x^2(3 - 2v)(v\,dx + x\,dv) = 0.$

   3. Divide through by $x^2$ (valid for $x \neq 0$)

   $(v^2 - 2v)\,dx + (3 - 2v)(v\,dx + x\,dv) = 0.$

   4. Collect the $dx$ and $dv$ terms

   $\big[(v^2 - 2v) + v(3 - 2v)\big]\,dx + (3 - 2v)x\,dv = 0.$

   Simplify the bracket:

   $(v^2 - 2v) + (3v - 2v^2) = -v^2 + v = v(1 - v).$

   So

   $v(1 - v)\,dx + (3 - 2v)x\,dv = 0.$

   5. Separate variables

   $\frac{dx}{x} = -\frac{3 - 2v}{v(1 - v)}\,dv.$

   6. Partial fractions Write $\dfrac{3 - 2v}{v(1 - v)} = \dfrac{A}{v} + \dfrac{B}{1 - v}$, so $3 - 2v = A(1 - v) + Bv$.
      • Set $v = 0$: $3 = A \Rightarrow A = 3$.
      • Set $v = 1$: $1 = B \Rightarrow B = 1$. Hence

   $\frac{3 - 2v}{v(1 - v)} = \frac{3}{v} + \frac{1}{1 - v}.$

   7. Integrate both sides $$$$

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4. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.The differential equation of the family of circles with fixed radius $r$ units and centre on the line $y = 3$, is (A) $1 + \left( \dfrac{dy}{dx} \right)^2 = \dfrac{r^2}{(y-3)^2}$ (B) $1 + \left( \dfrac{dy}{dx} \right)^2 = \dfrac{r^2}{y-3}$ (C) $\left( \dfrac{dy}{dx} \right)^2 = \dfrac{-r^2}{(y-3)^2}$ (D) $\left( \dfrac{dy}{dx} \right)^2 = \dfrac{r^2}{y-3}$

   ›Reveal solutionSolution

   The family of circles has centre $(h,3)$ and fixed radius $r$; eliminating the parameter $h$ by differentiating the circle equation and substituting back yields the differential equation $1 + (dy/dx)^2 = r^2/(y-3)^2$, which matches option (A).

   Concept & Intuition

   We have a family of curves (circles) that share a fixed radius $r$ and whose centres all lie on the horizontal line $y=3$. The only thing that varies from one circle to another is the $x$-coordinate of the centre, say $h$. To find the differential equation of the family, we need an equation that relates $x$, $y$, and $dy/dx$ without the parameter $h$. The standard approach: write the general equation of a circle in the family, differentiate it with respect to $x$, and then eliminate $h$ using the original equation.

   ---

   Step-by-step solution

   1. Write the equation of a typical circle A circle with centre $(h, 3)$ and radius $r$ has equation

   $(x - h)^2 + (y - 3)^2 = r^2.$

   Here $h$ is the parameter that distinguishes one circle from another.

   2. Differentiate implicitly with respect to $x$ Differentiating both sides:

   $2(x - h) + 2(y - 3) \frac{dy}{dx} = 0.$

   Divide through by 2:

   $(x - h) + (y - 3) \frac{dy}{dx} = 0.$

   This gives a relation between $x$, $y$, $h$, and $dy/dx$.

   3. Solve for $x - h$ From the differentiated equation:

   $x - h = - (y - 3) \frac{dy}{dx}.$

   4. Eliminate $h$ using the original circle equation Substitute $x - h$ from step 3 into the original circle equation:

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

   Simplify:

   $(y - 3)^2 \left( \frac{dy}{dx} \right)^2 + (y - 3)^2 = r^2.$

   5. Factor and rearrange Factor $(y - 3)^2$: $$$$

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5. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.If the order and degree of the differential equation corresponding to the family of curves $y^2 = 4a(x+a)$ ($a$ is parameter) are $m$ and $n$ respectively, then $m+n^2=$ (A) 3 (B) 4 (C) 5 (D) 2

   ›Reveal solutionSolution

   The differential equation for the family $y^2 = 4a(x+a)$ has order $m = 1$ and degree $n = 2$, so $m + n^2 = 1 + 4 = 5$. The correct option is (C).

   ---

   We start with the family of curves given by

   $y^2 = 4a(x + a),$

   where $a$ is a parameter. To find the differential equation, we eliminate $a$ by differentiating and then combining the equations.

   Concept & Intuition:

   A family of curves with one parameter yields a first-order differential equation. The order is the highest derivative present; the degree is the power of that highest derivative after the equation is made polynomial in derivatives. Here, because the parameter appears squared in the constant term, the elimination will produce a squared first derivative, giving degree 2.

   ---

   Step-by-step solution:

   1. Differentiate once with respect to $x$:

   $2y \frac{dy}{dx} = 4a \quad \Rightarrow \quad y y' = 2a,$

   where $y' = \frac{dy}{dx}$.

   2. Express $a$ in terms of $y$ and $y'$:

   $a = \frac{y y'}{2}.$

   3. Substitute back into the original equation to eliminate $a$:

   $y^2 = 4 \left( \frac{y y'}{2} \right) \left( x + \frac{y y'}{2} \right).$

   Simplify step by step:

   $y^2 = 2 y y' \left( x + \frac{y y'}{2} \right) = 2xy y' + y^2 (y')^2.$

   4. Rearrange to standard form:

   $y^2 = 2xy y' + y^2 (y')^2 \quad \Rightarrow \quad 0 = 2xy y' + y^2 (y')^2 - y^2.$

   Factor $y$ (assuming $y \neq 0$ for non-degenerate curves):

   $y \left[ 2x y' + y (y')^2 - y \right] = 0.$

   Since $y = 0$ gives only a trivial case, the differential equation is

   $2x y' + y (y')^2 - y = 0.$

   5. Determine order and degree: …

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6. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.The general solution of the differential equation $(x^2+2)dy+2xydx=e^{x^2+2}dx$ is (A) $\dfrac{x}{y}=e^{x^2-4}+c$ (B) $2xy=e^{x^2-2x+4}+c$ (C) $(x^2+2)y=e^{x^2-2x+4}+c$ (D) $(x^2+2)^2y=e^{x^2+2x-4}+c$

   ›Reveal solutionSolution

   The left-hand side is a perfect differential, $(x^2+2)\,dy+2xy\,dx=d\!\big[(x^2+2)y\big]$, so the integrating factor is $x^2+2$ and the solution has the form $(x^2+2)y=(\text{integral of the RHS})+c$ — option (C).

   Step 1 — put the equation in linear form.

   Dividing $(x^2+2)\,dy+2xy\,dx=e^{x^2+2}\,dx$ by $dx$:

   $(x^2+2)\frac{dy}{dx}+2xy=e^{x^2+2}\quad\Longrightarrow\quad \frac{dy}{dx}+\frac{2x}{x^2+2}\,y=\frac{e^{x^2+2}}{x^2+2}.$

   This is first-order linear with $P(x)=\dfrac{2x}{x^2+2}$.

   Step 2 — integrating factor.

   $\mu(x)=e^{\int \frac{2x}{x^2+2}\,dx}=e^{\log(x^2+2)}=x^2+2.$

   Step 3 — recognise the exact differential.

   Multiplying through by $\mu$ returns the original left side, and

   $\frac{d}{dx}\big[(x^2+2)y\big]=(x^2+2)\frac{dy}{dx}+2xy.$

   So the equation collapses to

   $d\big[(x^2+2)y\big]=(\text{right-hand side})\,dx,$

   and integrating gives the general solution in the form …

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7. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.If $m$ and $n$ are respectively the order and the degree of the differential equation representing the family of curves $y^2 - 5ax - 5a^{\frac{3}{2}} = 0$ ($a > 0$ is a parameter), then the value of $m - n$ is (A) $1$ (B) $-1$ (C) $2$ (D) $-2$

   ›Reveal solutionSolution

   The key idea is to eliminate the parameter $a$ from the given family to obtain the differential equation, then identify its order $m$ and degree $n$. The value of $m - n$ is $\boxed{-1}$.

   The problem gives a family of curves with a single parameter $a$. To find the differential equation that represents this family, we must eliminate $a$ by differentiating and then combining the equations. The order of the resulting differential equation is the highest derivative present, and the degree is the power of the highest derivative after the equation is made polynomial in derivatives.

   Let’s work through it step by step.

   1. Write the given family and differentiate once. The family is

   $y^2 - 5ax - 5a^{3/2} = 0, \quad a > 0.$

   Differentiate both sides with respect to $x$:

   $2y \frac{dy}{dx} - 5a = 0.$

   So

   $a = \frac{2y}{5} \frac{dy}{dx}.$

   2. Substitute $a$ back into the original equation. Replace $a$ in $y^2 - 5ax - 5a^{3/2} = 0$:

   $y^2 - 5\left(\frac{2y}{5} \frac{dy}{dx}\right)x - 5\left(\frac{2y}{5} \frac{dy}{dx}\right)^{3/2} = 0.$

   Simplify the second term:

   $y^2 - 2xy \frac{dy}{dx} - 5\left(\frac{2y}{5} \frac{dy}{dx}\right)^{3/2} = 0.$

   3. Isolate the term with the fractional exponent. Move the other terms to the other side:

   $5\left(\frac{2y}{5} \frac{dy}{dx}\right)^{3/2} = y^2 - 2xy \frac{dy}{dx}.$

   4. Square both sides to remove the $3/2$ exponent. This is necessary to get a polynomial in derivatives (so we can read the degree). Squaring:

   $25 \left(\frac{2y}{5} \frac{dy}{dx}\right)^3 = \left(y^2 - 2xy \frac{dy}{dx}\right)^2.$

   Simplify the left side:

   $25 \cdot \frac{8y^3}{125} \left(\frac{dy}{dx}\right)^3 = \frac{8y^3}{5} \left(\frac{dy}{dx}\right)^3.$ …

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8. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.The general solution of the differential equation $\dfrac{dy}{dx} + (\sec x \csc x) y = \cos^2 x$ is (A) $y \sec^2 x = \sin^2 x + c$ (B) $y \sec^2 x = \tan x + c$ (C) $y \tan x = \sin x \cos x + c$ (D) $2y \tan x = \sin^2 x + c$

   ›Reveal solutionSolution

   A linear ODE with integrating factor $\tan x$: it integrates to $y\tan x=\tfrac12\sin^2 x+c$, i.e. $2y\tan x=\sin^2 x+c$. Answer: (D).

   Linear form. $\dfrac{dy}{dx}+P(x)y=Q(x)$ with $P=\sec x\csc x=\dfrac{1}{\sin x\cos x}$ and $Q=\cos^2 x$.

   Integrating factor.

   $\int P\,dx=\int\frac{dx}{\sin x\cos x}=\int\frac{2\,dx}{\sin 2x}=\int 2\csc 2x\,dx=\log|\tan x|,$

   so $\mu=e^{\log|\tan x|}=\tan x$.

   Multiply through. The coefficient of $y$ becomes $\tan x\cdot\sec x\csc x=\sec^2 x$, so the left side is an exact derivative: …

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9. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.The general solution of the differential equation $\dfrac{dy}{dx} + (\sec x \csc x) y = \cos^2 x$ is (A) $y \sec^2 x = \tan x + c$ (B) $y \tan x = \sin x \cos x + c$ (C) $y \sec^2 x = \sin^2 x + c$ (D) $2y \tan x = \sin^2 x + c$

   ›Reveal solutionSolution

   $2y\tan x = \sin^2 x + c$ — option (D).

   This is a linear first-order ODE $\dfrac{dy}{dx} + P(x)\,y = Q(x)$ with $P = \sec x\csc x = \dfrac{1}{\sin x\cos x}$ and $Q = \cos^2 x$.

   Integrating factor:

   $\int \frac{dx}{\sin x\cos x} = \int \frac{\sec^2 x}{\tan x}\,dx = \log|\tan x| \;\Rightarrow\; \text{IF} = \tan x.$

   Multiply through and integrate: …

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10. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
    Q.The general solution of the differential equation $\dfrac{dy}{dx} = \dfrac{2x-3y+5}{6x-9y+7}$ is (A) $x-3y+\dfrac{22}{3}\log|3x-7|+c=0$ (B) $x-3y+\dfrac{8}{3}\log|6x-9y+1|+c=0$ (C) $3x-3y+\dfrac{8}{3}\log|3x-9y+1|+c=0$ (D) $3x-2y+\dfrac{22}{3}\log|2x-3y-7|+c=0$

    ›Reveal solutionSolution

    The equation is reducible to a homogeneous form by substituting $u = 2x - 3y$, because the coefficients of $x$ and $y$ in numerator and denominator are proportional. After separation and integration, the solution matches option (B).

    The key observation: the numerator $2x - 3y + 5$ and denominator $6x - 9y + 7$ have coefficients of $x$ and $y$ in the ratio $2:-3$ and $6:-9$ respectively — these ratios are equal ($\frac{2}{6} = \frac{-3}{-9} = \frac{1}{3}$). That means the equation is of the form $\frac{dy}{dx} = \frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y + c_2}$ where $\frac{a_1}{a_2} = \frac{b_1}{b_2}$. In such cases, the standard trick is to set $u = a_1 x + b_1 y$ (or any linear combination that simplifies both numerator and denominator to functions of $u$ alone). Here, the natural choice is $u = 2x - 3y$, because then $2x - 3y + 5 = u + 5$ and $6x - 9y + 7 = 3(2x - 3y) + 7 = 3u + 7$.

    1. Substitute $u = 2x - 3y$.

       Differentiate with respect to $x$:

       $\frac{du}{dx} = 2 - 3\frac{dy}{dx}$.

       Hence $\frac{dy}{dx} = \frac{2 - \frac{du}{dx}}{3}$.

    2. Rewrite the differential equation in terms of $u$.

       The original equation is

       $\frac{dy}{dx} = \frac{2x - 3y + 5}{6x - 9y + 7} = \frac{u + 5}{3u + 7}$.

       Substituting for $\frac{dy}{dx}$:

       $\frac{2 - \frac{du}{dx}}{3} = \frac{u + 5}{3u + 7}$.

    3. Solve for $\frac{du}{dx}$.

       Multiply both sides by 3:

       $2 - \frac{du}{dx} = \frac{3(u + 5)}{3u + 7}$.

       So

       $\frac{du}{dx} = 2 - \frac{3u + 15}{3u + 7}$.

       Combine the right-hand side over a common denominator:

       $\frac{du}{dx} = \frac{2(3u + 7) - (3u + 15)}{3u + 7} = \frac{6u + 14 - 3u - 15}{3u + 7} = \frac{3u - 1}{3u + 7}$.

    4. Separate variables and integrate.

       $\frac{du}{dx} = \frac{3u - 1}{3u + 7}$

       $\Rightarrow \frac{3u + 7}{3u - 1} \, du = dx$.

       Integrate both sides:

       $\int \frac{3u + 7}{3u - 1} \, du = \int dx$.

       To integrate the left side, perform polynomial division (or rewrite the numerator):

       $\frac{3u + 7}{3u - 1} = 1 + \frac{8}{3u - 1}$. …

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11. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
    Q.The general solution of the differential equation $dx = (2x + 3y - 4)dy$ is (A) $2x + 6y - 3\log|4x + 6y - 5| = c$ (B) $6y - 3\log|4x + 6y - 5| = c$ (C) $2x + 6y - 8 - 3\log|4x + 6y - 5| = c$ (D) $6x + 6y - 3\log|4x + 6y - 5| = c$

    ›Reveal solutionSolution

    The substitution $u = 2x + 3y$ turns the equation into a separable one; integrating and rearranging gives $6y - 3\log|4x + 6y - 5| = c$, option (B).

    The equation $dx = (2x + 3y - 4)\,dy$ means $\dfrac{dx}{dy} = 2x + 3y - 4$, i.e. it is linear in $x$ with $y$ as the independent variable. The cleanest route is a substitution.

    1. Substitute $u = 2x + 3y$. Then $\dfrac{du}{dy} = 2\dfrac{dx}{dy} + 3$. Since $\dfrac{dx}{dy} = 2x + 3y - 4 = u - 4$,

    $\frac{du}{dy} = 2(u - 4) + 3 = 2u - 5.$

    2. Separate and integrate.

    $\frac{du}{2u - 5} = dy \;\Rightarrow\; \frac{1}{2}\log|2u - 5| = y + C \;\Rightarrow\; \log|2u - 5| = 2y + C'.$

    3. Back-substitute $u = 2x + 3y$. $2u - 5 = 2(2x + 3y) - 5 = 4x + 6y - 5$, so

    $\log|4x + 6y - 5| = 2y + C'.$

    4. Rearrange to match the options. $2y - \log|4x + 6y - 5| = c_1$; multiplying through by $3$, …

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12. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.The substitution required to reduce the differential equation $t^2 dx + (x^2 - tx + t^2) dt = 0$ to a differential equation which can be solved by variables separable method is (A) $t = Vx$ (B) $ax + bt = Z$ (C) $V = tx^2$ (D) $x = tV^2$

    ›Reveal solutionSolution

    The given differential equation is homogeneous in $x$ and $t$ (each term has total degree 2), so the standard substitution $x = V t$ (or equivalently $t = V x$) reduces it to a separable equation. The correct choice is (A).

    We are given:

    $t^2\,dx + (x^2 - tx + t^2)\,dt = 0$

    Concept & Intuition

    A differential equation of the form $M(x,t)\,dx + N(x,t)\,dt = 0$ is homogeneous if $M$ and $N$ are homogeneous functions of the same degree. Here, each term in $t^2$, $x^2$, and $tx$ is degree 2. For such equations, the substitution $x = V t$ (or $t = V x$) makes the equation separable. Why? Because dividing numerator and denominator by the highest power of $t$ (or $x$) turns the equation into one involving only the ratio $x/t$ (or $t/x$), which then separates.

    Let’s work through it step by step.

    1. Check homogeneity Rewrite the equation as:

    $t^2\,dx + (x^2 - t x + t^2)\,dt = 0$

    The coefficient of $dx$ is $t^2$ (degree 2 in $t$ and $x$ if we treat $t$ and $x$ as variables). The coefficient of $dt$ is $x^2 - t x + t^2$: each term is degree 2. So the equation is homogeneous of degree 2.

    2. Choose the substitution

       For a homogeneous equation, the standard substitution is $x = V t$ (or $t = V x$). This replaces the two variables with one variable $V$ and the independent variable $t$ (or $x$). The option (A) says $t = Vx$, which is the same idea but swapping roles. Let’s test $x = V t$ first (the more common form), then see why (A) works.

    3. Apply $x = V t$

       Let $x = V t$, so $dx = V\,dt + t\,dV$. Substitute into the equation:

    $t^2 (V\,dt + t\,dV) + \big((V t)^2 - t(V t) + t^2\big)\,dt = 0$

    Simplify:

    $t^2 V\,dt + t^3\,dV + (V^2 t^2 - V t^2 + t^2)\,dt = 0$

    Factor $t^2$ from the $dt$ terms:

    $t^2 V\,dt + t^3\,dV + t^2(V^2 - V + 1)\,dt = 0$

    Combine the $dt$ terms:

    $t^2\big(V + V^2 - V + 1\big)dt + t^3\,dV = 0$

    Notice $V - V$ cancels, leaving:

    $t^2(V^2 + 1)\,dt + t^3\,dV = 0$

    4. Separate variables Divide through by $t^3 (V^2+1)$ (assuming $t \neq 0$): $$$$

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