Differential Equations – GUJCET Mathematics Study Notes
Target: Quick, clear, and exam‑friendly notes for the GUJCET Mathematics syllabus.
Scope: First‑order linear & separable equations, exact equations, second‑order linear equations with constant coefficients, and a touch of the variation‑of‑parameters method.
Format:
- Simple explanation
- Key points (bullet & table)
- Worked example (step‑by‑step)
- Quick revision summary
1. Simple Explanation
A differential equation connects an unknown function (y(x)) with its derivatives. It tells how the function changes, not what the function is.
Why it matters:
- Models real‑world phenomena (population growth, cooling, oscillations, circuits).
- Provides a systematic way to find (y(x)) that satisfies a given relation.
Typical form (for GUJCET):
[
\frac{dy}{dx} = f(x, y)
]
or for second order
[
\frac{d^2y}{dx^2} + p(x)\frac{dy}{dx} + q(x)y = r(x)
]
2. Key Points
| Topic | What to remember | Typical method |
|---|---|---|
| 1. Separable equations | (\displaystyle \frac{dy}{dx}=g(x)h(y)) | Separate (y) and (x) → integrate both sides |
| 2. Linear first‑order | (\displaystyle \frac{dy}{dx}+P(x)y=Q(x)) | Integrating factor (\mu(x)=e^{\int P(x),dx}) |
| 3. Exact equations | (M(x,y)+N(x,y),dy/dx=0) with (\partial M/\partial y=\partial N/\partial x) | Find potential function (\psi(x,y)) s.t. (\psi_x=M,\ \psi_y=N) |
| 4. Bernoulli | (\displaystyle \frac{dy}{dx}+P(x)y=Q(x)y^n) | Substitute (v=y^{1-n}) → linear in (v) |
| 5. Second‑order linear constant‑coefficient | (ay''+by'+cy=r(x)) | 1) Solve homogeneous: characteristic (ar^2+br+c=0); 2) Find particular via undetermined coefficients or variation of parameters |
| 6. Variation of parameters | For (y''+p(x)y'+q(x)y=r(x)) | Use solutions (y_1,y_2) of homogeneous part: (y_p=u_1y_1+u_2y_2) with (u_1',u_2') solving a 2×2 system |
| 7. Initial/Boundary conditions | Needed to find the specific solution (unique) | Plug into general solution to determine constants |
Quick Tips
- Check separability first – if you can write as (g(x)h(y)), you’re done in a few steps.
- For linear first‑order, compute (\mu(x)) once and reuse for similar problems.
- Exactness test: If (\partial M/\partial y \neq \partial N/\partial x), look for an integrating factor depending only on (x) or (y).
- Characteristic roots: Real distinct → two exponentials; repeated → (e^{rx}) and (xe^{rx}); complex → (e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)).
3. Worked Example
Problem (GUJCET style)
Solve the differential equation
[
\frac{dy}{dx} + 2y = x^2, \quad y(0)=1
]
Step‑by‑Step
-
Identify type
- Linear first‑order: (P(x)=2,\ Q(x)=x^2).
-
Find integrating factor
[ \mu(x)=e^{\int P(x),dx}=e^{\int 2,dx}=e^{2x} ] -
Multiply the equation by (\mu(x))
[ e^{2x}\frac{dy}{dx} + 2e^{2x}y = x^2e^{2x} ]Left side is (\dfrac{d}{dx}!\bigl(e^{2x}y\bigr)).
-
Integrate both sides
[ \frac{d}{dx}!\bigl(e^{2x}y\bigr) = x^2e^{2x} ] [ \int \frac{d}{dx}!\bigl(e^{2x}y\bigr),dx = \int x^2e^{2x},dx ] Use integration by parts twice (or a table).
Result:
[ e^{2x}y = \frac{e^{2x}}{2}\left(x^2 - x + \frac{1}{2}\right) + C ] -
Solve for (y)
[ y = \frac{1}{2}\left(x^2 - x + \frac{1}{2}\right) + Ce^{-2x} ] -
Apply initial condition (y(0)=1)
[ 1 = \frac{1}{2}\left(0 - 0 + \frac{1}{2}\right) + C \cdot 1 ] [ 1 = \frac{1}{4} + C \quad\Rightarrow\quad C = \frac{3}{4} ] -
Final solution