Q.Find the differential equation of system of concentric circles with centre $(1,\,2)$.

Showing the 12 most recent of 20 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.When the origin is shifted to the point $\left(-\dfrac{4}{7}, \dfrac{6}{7}\right)$ by translation of axes, if the transformed equation of $2x^2 + 5xy + 4y^2 - 2x - 4y + 2 = 0$ is $ax^2 + 35xy + by^2 + 2gx + 2fy + c = 0$, then (A) $a + b + c = 48$ (B) $2g + 2f + c = 28$ (C) $a + b = 2f + c$ (D) $a + c = 2g + b$

   ›Reveal solutionSolution

   Translating the origin to the given point eliminates the linear terms of the conic. Substituting and comparing coefficients (after clearing the fraction to match the given $35XY$ term) gives $a=14$, $b=28$, $c=6$, $g=f=0$. Checking each option, only (A) $a+b+c=48$ holds.

   We start with the original equation:

   $2x^2 + 5xy + 4y^2 - 2x - 4y + 2 = 0.$

   Shifting the origin to $\left(-\frac{4}{7}, \frac{6}{7}\right)$ means the old coordinates relate to the new ones $(X,Y)$ by:

   $x = X - \frac{4}{7}, \quad y = Y + \frac{6}{7}.$

   1. Substitute the translation into the original equation.

   Expanding each term:

   • $2x^2 = 2X^2 - \frac{16}{7}X + \frac{32}{49}$
   • $5xy = 5XY + \frac{30}{7}X - \frac{20}{7}Y - \frac{120}{49}$
   • $4y^2 = 4Y^2 + \frac{48}{7}Y + \frac{144}{49}$
   • $-2x = -2X + \frac{8}{7}$
   • $-4y = -4Y - \frac{24}{7}$
   • constant: $+2$

   2. Collect terms by $X^2$, $XY$, $Y^2$, $X$, $Y$, and constant.

   • $X^2$: $2X^2$
   • $XY$: $5XY$
   • $Y^2$: $4Y^2$
   • $X$ terms: $-\frac{16}{7}+\frac{30}{7}-\frac{14}{7}=0$
   • $Y$ terms: $-\frac{20}{7}+\frac{48}{7}-\frac{28}{7}=0$ (as expected, since the shift is to the conic's center, the linear terms vanish)
   • Constant: $\frac{32-120+144+56-168+98}{49} = \frac{42}{49} = \frac{6}{7}$

   So the transformed equation is:

   $2X^2 + 5XY + 4Y^2 + \frac{6}{7} = 0.$

   3. Match against the given form $aX^2+35XY+bY^2+2gX+2fY+c=0$.

   The given form has $35XY$, so multiply our equation by 7 to match:

   $14X^2 + 35XY + 28Y^2 + 6 = 0.$ …

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2. 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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3. 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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4. 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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5. 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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6. 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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7. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If $6x-5y-20=0$ is a normal to the ellipse $x^2+3y^2=k$, then $k=$ (A) $9$ (B) $17$ (C) $25$ (D) $37$

   ›Reveal solutionSolution

   The condition for a line to be normal to an ellipse leads to a relation between its slope and the ellipse’s parameters. Solving that relation gives $k = 37$, so the correct option is (D).

   ---

   The key idea: A line is normal to a curve if its slope is the negative reciprocal of the slope of the tangent at the point of contact. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the slope of the normal at $(x_1, y_1)$ is $\frac{a^2 y_1}{b^2 x_1}$. We match this to the given line’s slope and also enforce that the point lies on both the ellipse and the line.

   ---

   1. Rewrite the ellipse in standard form The ellipse is $x^2 + 3y^2 = k$. Divide through by $k$:

   $\frac{x^2}{k} + \frac{y^2}{k/3} = 1$

   So $a^2 = k$ and $b^2 = \frac{k}{3}$.

   2. Slope of the given normal line

      The line is $6x - 5y - 20 = 0$, or $y = \frac{6}{5}x - 4$. Its slope is $m = \frac{6}{5}$.

   3. Slope of the normal to the ellipse at a point $(x_1, y_1)$

      For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the slope of the normal is

   $m_{\text{normal}} = \frac{a^2 y_1}{b^2 x_1}$

   (derived from differentiating implicitly: $2x/a^2 + 2y y'/b^2 = 0$ gives tangent slope $-\frac{b^2 x}{a^2 y}$, so normal slope is its negative reciprocal).

   Substituting $a^2 = k$, $b^2 = k/3$:

   $m_{\text{normal}} = \frac{k \cdot y_1}{(k/3) \cdot x_1} = \frac{3y_1}{x_1}$

   4. Equate slopes Since the given line is normal,

   $\frac{3y_1}{x_1} = \frac{6}{5} \quad \Rightarrow \quad 15y_1 = 6x_1 \quad \Rightarrow \quad 5y_1 = 2x_1$

   So $y_1 = \frac{2}{5}x_1$.

   5. Point lies on the line The point $(x_1, y_1)$ also satisfies $6x_1 - 5y_1 - 20 = 0$. Substitute $y_1 = \frac{2}{5}x_1$: $$$$

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8. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If $y = \sin ax + \cos bx$ then $y'' + b^2 y =$ (A) $(b^2 - a^2) \sin ax$ (B) $(b^2 - a^2) \cos bx$ (C) $(a^2 - b^2) \tan ax$ (D) $(b^2 - a^2) \cot bx$

   ›Reveal solutionSolution

   The key idea is to compute the second derivative of $y = \sin ax + \cos bx$ and then substitute into $y'' + b^2 y$. The cross terms cancel, leaving only $(b^2 - a^2)\sin ax$, so the correct option is (A).

   We start with the function

   $y = \sin(ax) + \cos(bx).$

   The problem asks for $y'' + b^2 y$. The natural approach is to differentiate twice and then combine terms. Notice that the second derivative of $\sin(ax)$ will bring down a factor of $-a^2$, and the second derivative of $\cos(bx)$ will bring down a factor of $-b^2$. When we add $b^2 y$, the $\cos(bx)$ part will cancel, leaving only a term involving $\sin(ax)$. This is the core insight.

   Let’s work through it step by step.

   1. First derivative Differentiate each term:

   $y' = a\cos(ax) - b\sin(bx).$

   (Derivative of $\sin(ax)$ is $a\cos(ax)$; derivative of $\cos(bx)$ is $-b\sin(bx)$.)

   2. Second derivative Differentiate again:

   $y'' = -a^2\sin(ax) - b^2\cos(bx).$

   (Derivative of $a\cos(ax)$ is $-a^2\sin(ax)$; derivative of $-b\sin(bx)$ is $-b^2\cos(bx)$.)

   3. Form the expression $y'' + b^2 y$ Substitute $y''$ and $y$:

   $y'' + b^2 y = \bigl[-a^2\sin(ax) - b^2\cos(bx)\bigr] + b^2\bigl[\sin(ax) + \cos(bx)\bigr].$

   4. Simplify Group the $\sin(ax)$ terms and the $\cos(bx)$ terms: $$$$

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9. 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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10. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.$(0, k)$ is the point to which the origin is to be shifted by the translation of the axes so as to remove the first degree terms from the equation $ax^2 - 2xy + by^2 - 2x + 4y + 1 = 0$ and $\dfrac{1}{2} \tan^{-1}(2)$ is the angle through which the coordinate axes are to be rotated about the origin to remove the $xy$-term from the given equation, then $a + b =$ (A) $1$ (B) $-2$ (C) $3$ (D) $-4$

    ›Reveal solutionSolution

    The problem combines translation to eliminate linear terms and rotation to eliminate the $xy$-term; solving the conditions yields $a+b = 3$, so the correct option is (C).

    We are given the equation

    $ax^2 - 2xy + by^2 - 2x + 4y + 1 = 0.$

    We shift the origin to $(0,k)$ to remove the first-degree terms, and we rotate the axes by $\frac12 \tan^{-1}(2)$ to remove the $xy$-term. We need $a+b$.

    ---

    Concept and intuition

    When we translate axes, the linear terms in the new equation come from the gradient of the quadratic part evaluated at the new origin. Setting them to zero gives conditions on $a,b$ and the shift.

    When we rotate axes, the coefficient of $xy$ in the rotated equation is zero if the rotation angle $\theta$ satisfies $\cot 2\theta = \frac{a-b}{-2}$ (for a general $Ax^2 + Bxy + Cy^2$). Here we are given $\theta$, so we can find a relation between $a$ and $b$.

    ---

    Step-by-step solution

    1. Translation to remove first-degree terms

    Let the new origin be $(0,k)$. Substitute $x = X$, $y = Y + k$ into the equation:

    $aX^2 - 2X(Y+k) + b(Y+k)^2 - 2X + 4(Y+k) + 1 = 0.$

    Expand:

    $aX^2 - 2XY - 2kX + b(Y^2 + 2kY + k^2) - 2X + 4Y + 4k + 1 = 0.$

    Collect terms in $X$ and $Y$:

    • Coefficient of $X$: $-2k - 2$
    • Coefficient of $Y$: $2bk + 4$

    For the first-degree terms to vanish, we set these to zero:

    $-2k - 2 = 0 \quad\Rightarrow\quad k = -1,$

    $2bk + 4 = 0 \quad\Rightarrow\quad 2b(-1) + 4 = 0 \quad\Rightarrow\quad -2b + 4 = 0 \quad\Rightarrow\quad b = 2.$

    So $b = 2$ is determined from translation alone.

    2. Rotation to remove the $xy$-term

    The general second-degree equation $Ax^2 + Bxy + Cy^2 + \dots$ has $B = -2$ here. To eliminate the $xy$-term, we rotate by angle $\theta$ satisfying

    $\cot 2\theta = \frac{A - C}{B}.$

    Here $A = a$, $C = b = 2$, $B = -2$. So

    $$$$

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11. 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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12. 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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