Jamia Millia Islamia MCA Differential Equation Previous Year Questions (PYQs) – Page 1 of 2

**Solution:** To find the degree, remove the fractional exponent by squaring both sides: $\left[\,1 + \left(\dfrac{dy}{dx}\right)^2\right]^3 = \left(\dfrac{d^2y}{dx^2}\right)^2$ Now the equation is polynomial in derivatives, and the highest order derivative is $\dfrac{d^2y}{dx^2}$ appearing as a square term. Therefore, **Degree = 2** $\boxed{\text{Answer: (D) 2}}$

Given: $\dfrac{dy}{dx} = e^{x - y} + x^2 e^{-y} = e^{-y}(e^x + x^2)$ Multiply both sides by $e^y$: $e^y \dfrac{dy}{dx} = e^x + x^2$ Integrate both sides: $\int e^y dy = \int (e^x + x^2)\,dx$ $\Rightarrow e^y = e^x + \dfrac{x^3}{3} + c$ $\boxed{\text{Answer: (B)}}$

Let $x$ be a function of $y.$ $\frac{dx}{dy} = \frac{\tan^{-1}y - x}{1 + y^2}$ $\Rightarrow \frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{-1}y}{1 + y^2}$ This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y),$ where $P(y) = \frac{1}{1 + y^2}.$ Integrating factor, $\mu = e^{\int P(y) dy} = e^{\tan^{-1}y}.$ $\Rightarrow \frac{d}{dy}(xe^{\tan^{-1}y}) = e^{\tan^{-1}y} \frac{\tan^{-1}y}{1 + y^2}$ Let $t = \tan^{-1}y \Rightarrow dt = \frac{dy}{1 + y^2}$ $xe^{t} = \int t e^{t} dt = e^{t}(t - 1) + C$ $\Rightarrow x = \tan^{-1}y - 1 + Ce^{-\tan^{-1}y}$

Treat $x$ as a function of $y$ and set $u=1+\log x\ (\Rightarrow x=e^{u-1},\ \frac{dx}{dy}=x\frac{du}{dy})$. The DE becomes $u\frac{du}{dy}=u-1+e^{\,1+y-u}.$ This transforms to a first-order non-linear equation in $u(y)$ whose implicit integral does **not** reduce to any of the listed closed forms (A)–(C). With the initial condition $y(1)=0$ (i.e., $u=1$ at $y=0$), the solution is an implicit relation not matching (A)–(C).

The given equation is $2\dfrac{dy}{dx} + x^2 y = 2x + 3$. Dividing by 2: $\dfrac{dy}{dx} + \dfrac{x^2}{2}y = x + \dfrac{3}{2}$. This is a linear differential equation in $y$ with fixed constants.

Given $y = c(x - c)^2 = c(x^2 - 2cx + c^2) = c x^2 - 2c^2 x + c^3$. Differentiate three times to eliminate $c$. Hence, the differential equation will be of order 3.

For $y = ae^{-x}$, $\dfrac{dy}{dx} = -ae^{-x} = -y$. For $y = be^{x}$, $\dfrac{dy}{dx} = be^{x} = y$. At point of intersection: $ae^{-x} = be^{x} \Rightarrow e^{2x} = \dfrac{a}{b}$. Slopes: $m_1 = -\dfrac{b}{a}$ and $m_2 = \dfrac{a}{b}$. For orthogonality: $m_1 m_2 = -1 \Rightarrow ab = 1$.

$\dfrac{dy}{dx}=-\dfrac{1}{x}\;\Rightarrow\; y=-\ln|x|+C$. Using $y(-1)=0$ gives $C=0$. Hence $y=-\ln|x|$ (same as $-\log|x|$ up to base).

The order of a differential equation is the highest derivative present. Here, the highest derivative is $y''$ $\Rightarrow$ Order = 2.