MAH-CET 2026 — Mathematics PYQ

1. Identify the Type of Differential Equation:

The given differential equation is:

$$\frac{dy}{dx}=\frac{y^2-2xy-x^2}{y^2+2xy-x^2}$$

Since every term in the numerator and denominator has a combined degree of 2, this is a homogeneous differential equation.

2. Substitution for Homogeneous Equations:

To solve it, substitute $y=vx$.

Differentiating both sides with respect to $x$ using the product rule gives:

$$\frac{dy}{dx}=v+x\frac{dv}{dx}$$

Substitute $y=vx$ and $\frac{dy}{dx}$ back into the original differential equation:

$$v+x\frac{dv}{dx}=\frac{(vx{)}^2-2x(vx)-x^2}{(vx{)}^2+2x(vx)-x^2}$$

$$v+x\frac{dv}{dx}=\frac{x^2(v^2-2v-1)}{x^2(v^2+2v-1)}$$

Canceling out $x^2$ from the numerator and denominator:

$$v+x\frac{dv}{dx}=\frac{v^2-2v-1}{v^2+2v-1}$$

3. Separate the Variables:

Move $v$ to the right-hand side:

$$x\frac{dv}{dx}=\frac{v^2-2v-1}{v^2+2v-1}-v$$

$$x\frac{dv}{dx}=\frac{(v^2-2v-1)-v(v^2+2v-1)}{v^2+2v-1}$$

$$x\frac{dv}{dx}=\frac{v^2-2v-1-v^3-2v^2+v}{v^2+2v-1}$$

$$x\frac{dv}{dx}=\frac{-v^3-v^2-v-1}{v^2+2v-1}$$

$$x\frac{dv}{dx}=-\frac{v^3+v^2+v+1}{v^2+2v-1}$$

Factorize the numerator: $v^3+v^2+v+1=v^2(v+1)+1(v+1)=(v^2+1)(v+1)$.

$$x\frac{dv}{dx}=-\frac{(v^2+1)(v+1)}{v^2+2v-1}$$

Rearranging the terms to separate variables $v$ and $x$:

$$\frac{v^2+2v-1}{(v^2+1)(v+1)}dv=-\frac{dx}{x}$$

4. Integrate Using Partial Fractions:

Let's resolve the left-hand side into partial fractions:

$$\frac{v^2+2v-1}{(v^2+1)(v+1)}=\frac{Av+B}{v^2+1}+\frac{C}{v+1}$$

Multiplying out the denominators:

$$v^2+2v-1=(Av+B)(v+1)+C(v^2+1)$$

$$v^2+2v-1=A{v}^2+Av+Bv+B+C{v}^2+C$$

$$v^2+2v-1=(A+C)v^2+(A+B)v+(B+C)$$

Comparing the coefficients on both sides:

1. $A+C=1$

2. $A+B=2$

3. $B+C=-1$

Subtracting equation (3) from equation (2) gives:

$$(A+B)-(B+C)=2-(-1)⟹A-C=3$$

Adding this to equation (1) ($A+C=1$):

$$2A=4⟹A=2$$

Now find $B$ and $C$:

• From $A+C=1⟹2+C=1⟹C=-1$

• From $A+B=2⟹2+B=2⟹B=0$

Thus, the partial fractions expression is:

$$\frac{2v}{v^2+1}-\frac{1}{v+1}$$

Now, integrate both sides:

$$\int \left(\frac{2v}{v^2+1}-\frac{1}{v+1}\right)dv=-\int \frac{dx}{x}$$

$$\ln (v^2+1)-\ln (v+1)=-\ln |x|+\ln |c|$$

Using logarithm properties ($\ln a-\ln b=\ln (\frac{a}{b})$):

$$\ln \left(\frac{v^2+1}{v+1}\right)=\ln \left(\frac{c}{x}\right)$$

$$\frac{v^2+1}{v+1}=\frac{c}{x}$$

5. Substitute back $v=\frac{y}{x}$:

$$\frac{{\left(\frac{y}{x}\right)}^2+1}{\frac{y}{x}+1}=\frac{c}{x}$$

$$\frac{\frac{y^2+x^2}{x^2}}{\frac{y+x}{x}}=\frac{c}{x}$$

$$\frac{x^2+y^2}{x(x+y)}=\frac{c}{x}$$

$$x^2+y^2=c(x+y)$$

6. Apply the Initial Condition:

We are given that $y(-1)=1$, which means when $x=-1$, $y=1$:

$$(-1{)}^2+(1{)}^2=c(-1+1)$$

$$1+1=c(0)$$

$$2=0⟹\text{No finite solution for }c\text{ under this grouping form.}$$

Let's re-verify the integration constant layout. If we use the constant as an additive term:

$$\ln (v^2+1)-\ln (v+1)+\ln |x|=\ln |c|$$

$$\ln \left|\frac{x(v^2+1)}{v+1}\right|=\ln |c|⟹\frac{x\left(\frac{x^2+y^2}{x^2}\right)}{\frac{x+y}{x}}=c⟹x^2+y^2=c(x+y)$$

Let's re-evaluate the original differential equation directly by cross-multiplying to spot the shortcut form:

$$(y^2+2xy-x^2)dy=(y^2-2xy-x^2)dx$$

$$(y^2-x^2)dy+2xydy=(y^2-x^2)dx-2xydx$$

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

Alternatively, group the standard implicit forms:

$$x^{2}dx+2xydx-y^{2}dx+y^{2}dy-2xydy-x^{2}dy=0$$

$$(x^2-y^2)dx+2xydx-(x^2-y^2)dy-2xydy=0$$

Let's inspect the general second-degree algebraic form of the result:

$$x^2+y^2-cx-cy=0$$

An equation of the form $x^2+y^2+2gx+2fy+c^{\prime }=0$ always represents a circle because:

• The coefficients of $x^2$ and $y^2$ are equal.

• There is no $xy$ term present.

Therefore, the solution curve is a circle.