NDA (National Defence Academy) · Mathematics (JEE & NDA)
Differential Equations
Order and degree of differential equations, separation of variables, homogeneous equations and linear first order equations.
Six concepts from order and degree through separable, homogeneous and linear first-order equations to growth models — each one the type-match that tells you which integral to write.
- NDA (National Defence Academy)
- Medium level
- 6 concepts
- 5 practice questions
1Order, degree and formation
The order of a differential equation is the highest derivative present; the degree is the power of that highest derivative once the equation is polynomial in the derivatives. A family carrying n arbitrary constants yields an n-th order equation when those constants are eliminated — so order tracks how many free constants the general solution still has.
Hold the equation and ask two questions — which is the highest derivative, and to what power does it appear once the equation is polynomial in derivatives.
| Feature | What to look at | Example |
|---|---|---|
| Order | Highest derivative | y''+y=0 has order 2 |
| Degree | Power of that highest derivative after clearing roots | (y')^2=y has degree 2 |
| Formation | Eliminate n constants from a family | y=Ae^{kx} yields a first-order equation in y |
The equation (y'')^3 + (y')^4 + y = 0 has
- order 2, degree 3
- order 3, degree 2
- order 2, degree 4
The highest derivative is y'' (order 2), raised to the third power (degree 3). The fourth power sits on a lower derivative and does not set the degree.
2Variable-separable equations
If \dfrac{dy}{dx}=f(x)g(y) and g(y)\neq 0, separate and integrate: \int\dfrac{dy}{g(y)}=\int f(x)\,dx+C. The constant is fixed only after the indefinite integrals are done — fold it into a single C, then apply the initial condition.
Figure. After separating \mathrm{d}y/\mathrm{d}x=x/y, the solutions are y^2-x^2=C. Different C are different curves; (0,1) forces C=1.
Separating variables
- Factor the right-hand sideWrite \dfrac{dy}{dx} as a product f(x)g(y) (or rearrange until it is one).
- Separate and integrateMove every y factor with dy and every x factor with dx, then integrate both sides.
- Fix C lastApply the initial condition only after the antiderivatives are written and the constant is collected.
Separable equation
Solve \dfrac{dy}{dx}=\dfrac{x}{y} with y(0)=3.
- Separate: y\,dy=x\,dxvariables apart
- \int y\,dy=\int x\,dx\dfrac{y^2}{2}=\dfrac{x^2}{2}+C
- y(0)=3 gives \dfrac{9}{2}=C, so y^2-x^2=2Cy^2-x^2=9
Pro tip. Fold the constant of integration into a single constant and fix it with the initial condition at the very end.
The solution of \dfrac{dy}{dx}=\dfrac{x}{y} with y(0)=3 is
- y^2-x^2=9
- y^2+x^2=9
- y=3x
Separation gives y\,dy=x\,dx; integrating and using y(0)=3 yields the hyperbola y^2-x^2=9.
3Homogeneous equations
If \dfrac{dy}{dx}=F\!\left(\dfrac{y}{x}\right), the right-hand side depends on the ratio alone. Substitute y=vx so \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}; the equation collapses to a separable one in v and x. After integrating, replace v by y/x.
Figure. A first-order equation whose right-hand side is a function of y/x alone takes y=vx. The chain rule replaces dy/dx by v+x\,dv/dx, and the equation becomes separable in v and x.
The y=vx reduction
- Confirm the ratio formCheck that replacing (x,y) by (kx,ky) leaves \dfrac{dy}{dx} unchanged — equivalently, that it is a function of y/x only.
- Substitute y=vxWrite \dfrac{dy}{dx}=v+x\dfrac{dv}{dx} and replace every y/x by v.
- Separate in v,xSolve the separable equation for v(x), then restore y=vx.
For a homogeneous first-order ODE, the standard substitution is
- y=vx, giving \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}
- x=vy
- the integrating factor e^{\int P\,dx}
Homogeneous means F(y/x); y=vx produces the separable pair in v and x. The IF is the linear-equation tool, not this one.
4Linear first-order and the integrating factor
For \dfrac{dy}{dx}+P(x)y=Q(x), the integrating factor is \mu=e^{\int P\,dx}. Multiply through and the left side becomes exactly \dfrac{d}{dx}(y\mu), so the solution is y\,\mu=\int Q\,\mu\,dx+C. Recognising that exact derivative saves re-deriving the product rule on every problem.
Figure. For y'+y/x=x the integrating factor is \mu=e^{\int dx/x}=x. Multiplying through makes the left side d(xy)/dx=x^2, so xy=\int x^2\,dx and y=x^2/3+C/x.
Solving a linear first-order equation
- Write standard formArrange as y'+P(x)y=Q(x) with the coefficient of y' equal to 1.
- Build \muCompute \mu=e^{\int P\,dx} (drop the constant inside the exponential — it cancels).
- Integrate the right sideMultiply through by \mu, replace the left by \dfrac{d}{dx}(y\mu), and integrate Q\mu.
Linear first-order equation
Solve \dfrac{dy}{dx}+\dfrac{y}{x}=x.
- P=\dfrac{1}{x}, so \mu=e^{\int dx/x}\mu=e^{\ln x}=x
- \dfrac{d}{dx}(xy)=x\cdot xx^2
- xy=\int x^2\,dxy=\dfrac{x^2}{3}+\dfrac{C}{x}
Pro tip. After multiplying by \mu, the left side is exactly \dfrac{d}{dx}(y\mu); recognising this saves re-deriving the product rule.
For y'+\dfrac{y}{x}=x, the integrating factor is
- x
- e^{x}
- 1/x
P=1/x gives \mu=e^{\int dx/x}=x (for x>0).
5Identify the type before integrating
Matching the form immediately fixes the method: separable when f(x)g(y), homogeneous when F(y/x), linear when y'+Py=Q. Starting the wrong integral wastes the paper; the first thirty seconds of a DE question are a classification, not an integration.
Three shapes, three first moves — separable, homogeneous, linear — decided before any integral is written.
| Look of dy/dx | Type | First move |
|---|---|---|
| f(x)g(y) | Separable | Separate and integrate |
| F(y/x) | Homogeneous | Set y=vx |
| y'+P(x)y=Q(x) | Linear | Build \mu=e^{\int P\,dx} |
The equation \dfrac{dy}{dx}=\dfrac{x+y}{x-y} is first treated as
- homogeneous, via y=vx
- linear in y, via an IF
- separable as written
Dividing numerator and denominator by x shows a function of y/x only, so the homogeneous substitution applies.
6Growth, decay and linear models
Growth and decay follow \dfrac{dN}{dt}=kN, giving N=N_0e^{kt}. Newton's law of cooling and simple mixing problems reduce to first-order linear equations once the rate is written as proportional to a difference — so the IF method from the linear card finishes them.
Figure. Exponential growth from N_0 at t=0: the series is N=N_0e^{kt} sampled in the plot frame, starting at the origin of the drawn axes.
Reading a growth model
- Write the ratePure growth/decay is \dfrac{dN}{dt}=kN; cooling/mixing is usually linear in the unknown.
- Solve the DESeparate for kN, or build an IF for the linear form.
- Fit the dataUse a given value (or doubling time) to fix N_0 and k.
Exponential growth from dN/dt = kN
If \dfrac{dN}{dt}=kN and N(0)=N_0, show that N=N_0e^{kt}.
- Separate: \dfrac{dN}{N}=k\,dtvariables apart
- \int\dfrac{dN}{N}=\int k\,dt\ln|N|=kt+C
- N(0)=N_0 fixes e^{C}=N_0N=N_0e^{kt}
Pro tip. The same separation with k<0 is exponential decay; only the sign of k changes.
The DE \dfrac{dN}{dt}=kN has solutions of the form
- N=N_0e^{kt}
- N=N_0+kt
- N=k/N_0
Separating and integrating \mathrm{d}N/N=k\,\mathrm{d}t produces the exponential family N=N_0e^{kt}.
Notes
- Order, degree and formation: The order of a differential equation is the highest derivative present; the degree is the power of that highest derivative once the equation is polynomial in derivatives. A family with n arbitrary constants yields an n-th order equation on elimination.
- Variable-separable equations: If \dfrac{dy}{dx}=f(x)g(y), separate and integrate: \int\dfrac{dy}{g(y)}=\int f(x)\,dx+C.
- Homogeneous equations: If \dfrac{dy}{dx}=F\!\left(\dfrac{y}{x}\right), substitute y=vx so \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}, reducing it to a variable-separable equation in v and x.
- Linear first-order equations: For \dfrac{dy}{dx}+P(x)y=Q(x), the integrating factor is \mu=e^{\int P\,dx}, and the solution is y\,\mu=\int Q\,\mu\,dx+C.
- Applications: Growth and decay follow \dfrac{dN}{dt}=kN giving N=N_0e^{kt}; Newton's law of cooling and simple mixing problems reduce to first-order linear equations.
Formulas
- Separable: \int\dfrac{dy}{g(y)}=\int f(x)\,dx+C
- Homogeneous: substitute y=vx,\ \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}
- Linear form: \dfrac{dy}{dx}+P(x)y=Q(x)
- Integrating factor: \mu=e^{\int P\,dx}
- Linear solution: y\,\mu=\int Q\,\mu\,dx+C
- Growth/decay: \dfrac{dN}{dt}=kN\Rightarrow N=N_0e^{kt}
Exam traps & shortcuts
- Identify the type first: separable, homogeneous or linear - matching the form immediately fixes the method and avoids wasted algebra.
- For a linear equation, once you have \mu=e^{\int P\,dx}, the left side is always the exact derivative \dfrac{d}{dx}(y\mu), so you only integrate the right side.
- To check a proposed solution, differentiate it and substitute back rather than re-solving - fast for MCQs with given options.
Reference tables
Match the shape, then open the matching integral.
| Shape | Tool |
|---|---|
| \dfrac{dy}{dx}=f(x)g(y) | \int dy/g=\int f\,dx+C |
| \dfrac{dy}{dx}=F(y/x) | y=vx, then separable in v,x |
| y'+P(x)y=Q(x) | \mu=e^{\int P\,dx}, then y\mu=\int Q\mu\,dx+C |
| \dfrac{dN}{dt}=kN | N=N_0e^{kt} |
Recap
Read only this the night before.
- Order / degree
- Order = highest derivative; degree = its power once the equation is polynomial in derivatives.
- Separable
- Split, integrate, fix C last with the initial condition.
- Homogeneous
- F(y/x) then y=vx, then separable in v and x.
- Linear
- \mu=e^{\int P\,dx} turns the left side into \dfrac{d}{dx}(y\mu).
- Type first
- Classify before integrating — wrong method wastes the page.
- Growth
- \dfrac{dN}{dt}=kN\Rightarrow N=N_0e^{kt}; cooling/mixing uses the linear IF.
Practise Differential Equations
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- 5 exam-style questions on this topic, with explanations
- A 6-question practice set that ends the chapter
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