Solution for CBSE Class 12 Mathematics (Differential Equations)

(a) Find the particular solution of the differential equation:

$$\frac{dy}{dx} - 2xy = 3x^2 e^{x^2}; \quad y(0) = 5$$

Step 1 (Formula / Governing Law & Given Data): [1 Mark]

The given equation is a first-order linear differential equation of the form:

$$\frac{dy}{dx} + P(x)y = Q(x)$$

Comparing the given equation, we identify:

• $P(x) = -2x$
• $Q(x) = 3x^2 e^{x^2}$

The Integrating Factor (I.F.) is calculated as $I.F. = e^{\int P(x) dx}$.

Step 2 (Step-by-Step Mathematical Working): [3 Marks]

Calculate the Integrating Factor:

$$I.F. = e^{\int -2x \, dx} = e^{-x^2}$$

The general solution is given by $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \, dx + C$:

$$y \cdot e^{-x^2} = \int (3x^2 e^{x^2}) \cdot (e^{-x^2}) \, dx + C$$

$$y \cdot e^{-x^2} = \int 3x^2 \, dx + C$$

$$y \cdot e^{-x^2} = x^3 + C$$

Now, apply the initial condition $y(0) = 5$ to find $C$:

$$5 \cdot e^{-(0)^2} = (0)^3 + C \implies 5 \cdot 1 = 0 + C \implies C = 5$$

Step 3 (Final Concluded Answer): [1 Mark]

Substituting $C = 5$ into the general solution:

$$y \cdot e^{-x^2} = x^3 + 5$$

The particular solution is:

$$y = (x^3 + 5)e^{x^2}$$

💡 EL10 Concept Insight:

Think of the Integrating Factor as a "magic multiplier" that turns the messy left side of the equation into a simple product rule derivative, making it easy to undo the differentiation by integrating both sides.