Questions: Differential Equations

Q1 Find the order and degree of the differential equation obtained by squaring d squared y by dx squared plus dy by dx equals y.

Answer (English): The equation is second order because the highest derivative is the second derivative. After squaring, the highest derivative appears squared, so the degree is two. The squared equation is not polynomial in the derivatives at first sight because of the mixed term, but after expansion both sides are polynomials in the derivatives and the degree is read from the power of d squared y by dx squared, which is two.

Hinglish: Highest derivative second derivative hai, isliye order 2. Square karne ke baad uski power 2 ho jaati hai, isliye degree 2. Radicals ya fractions ho to pehle clear karo, tabhi degree batao.

Q2 Form the differential equation of the family of circles centred at the origin.

Answer (English): The family is x squared plus y squared equal to a squared, with one parameter a. Differentiate once with respect to x to get two x plus two y dy by dx equal zero. The parameter has already disappeared, so the differential equation is x plus y dy by dx equal zero, which is first order. The order matches the number of parameters, as expected.

Hinglish: Family x² + y² = a² hai, ek parameter. Ek baar differentiate karo: 2x + 2y y' = 0, yani x + y y' = 0. Parameter khud nikal gaya, aur order 1 hai jo parameters ki ginti se match karta hai.

Q3 Solve dy by dx equal to y over x, given that y equals two when x equals one.

Answer (English): Separate the variables to get dy by y equal dx by x, and integrate to log y equal log x plus a constant. Exponentiating gives y equal C x, with the absolute values absorbed into the constant. Using the condition gives two equal C times one, so C is two and the particular solution is y equal two x. The solution is a line through the origin, consistent with the family of proportional relations.

Hinglish: Variables alag karke ∫dy/y = ∫dx/x likho, to log y = log x + C, yani y = Cx. Condition se C = 2, isliye y = 2x particular solution hai.

Q4 Solve dy by dx equal to x squared y cubed with y equal to one when x equals zero.

Answer (English): Separate to y to the power minus three dy equal x squared dx. Integrate to get minus one half y to the minus two equal x cubed over three plus C. Using y of zero equal to one gives minus one half equal C, so the particular solution is minus one half y to the minus two equal x cubed over three minus one half. Multiplying by minus two and rearranging gives y to the minus two equal one minus two x cubed over three, or y squared equal three over three minus two x cubed.

Hinglish: y^{-3} dy = x² dx likho aur integrate karo: −(1/2)y^{-2} = x³/3 + C. y(0) = 1 se C = −1/2, isliye answer y² = 3/(3 − 2x³) aata hai.

Q5 Solve the homogeneous equation x dy minus y dx equal to zero and interpret the solution.

Answer (English): Rewrite as dy by dx equal y over x, which is already in the homogeneous form. Substituting y equal to v x gives v plus x dv by dx equal v, so x dv by dx is zero and v is constant. Therefore y equal C x, a family of straight lines through the origin. The result also follows from separation; the substitution confirms that a homogeneous equation of this simple type collapses immediately. The lines are exactly the level curves of the ratio y over x.

Hinglish: Equation dy/dx = y/x homogeneous hai. y = vx substitute karo to v + x dv/dx = v, isliye x dv/dx = 0 aur v constant. Answer y = Cx, yani origin se guzarne wali lines ka family.

Q6 Solve dy by dx plus y over x equal x squared.

Answer (English): The equation is linear with P equal one over x and Q equal x squared. The integrating factor is e to the integral of one over x, which is x. Multiplying the equation by x gives the derivative of x y equal to x cubed. Integrating gives x y equal x to the fourth over four plus C, so y equal x cubed over four plus C over x. Differentiating this answer returns x squared over four plus x squared over four minus the constant term, which simplifies to the original left side.

Hinglish: Linear form mein P = 1/x aur Q = x². IF = x milta hai. Multiply karke (xy)' = x³ likho, integrate karke xy = x⁴/4 + C, isliye y = x³/4 + C/x.

Q7 Solve dy by dx plus y cot x equal cosine squared x, given that y equals zero when x equals pi by two.

Answer (English): The equation is linear with P equal cotangent x, so the integrating factor is e to the integral of cotangent x, which is sine x. Multiplying gives the derivative of y sine x equal to sine x cosine squared x. The integral of the right side is minus cosine cubed x over three, so y sine x equal minus cosine cubed x over three plus C. The initial condition at pi by two gives zero equal C, so the particular solution is y equal minus cosine cubed x over three sine x.

Hinglish: IF = sin x. Multiply karke (y sinx)' = sinx cos²x. Integrate karke y sinx = −cos³x/3 + C. x = π/2 par y = 0 se C = 0, isliye y = −cos³x/(3 sinx).

Q8 A population grows at a rate proportional to its size. If the population doubles in ten years, find the time needed to triple.

Answer (English): Let the population be P at time t. The statement gives dP by dt equal k P, which is separable. Integrate to log P equal k t plus a constant, so P equal P nought e to the k t. The doubling condition gives two P nought equal P nought e to the ten k, so k equals the logarithm of two divided by ten. The tripling condition is three P nought equal P nought e to the k t, so t equals the logarithm of three divided by k, which is ten times the logarithm of three divided by the logarithm of two, about fifteen point eight years.

Hinglish: dP/dt = kP se P = P₀e^{kt}. Doubling se 2 = e^{10k}, yani k = ln2/10. Tripling ke liye 3 = e^{kt}, isliye t = 10 ln3/ln2 ≈ 15.8 saal.

Q9 Find the particular solution of dy by dx plus two y tangent x equal sin x, given that y equals zero when x equals pi by three.

Answer (English): The equation is linear with P equal two tangent x, and the integrating factor is e to the integral of two tangent x, which is secant squared x. Multiplying gives the derivative of y secant squared x equal to sine x over cosine squared x, whose integral is secant x. So y secant squared x equal secant x plus C. The condition at pi by three gives zero equal two plus C, so C is minus two and the solution is y equal cosine x minus two cosine squared x.

Hinglish: IF = sec²x. Multiply karke (y sec²x)' = sinx/cos²x, jiska integral secx hai. Isliye y sec²x = secx + C. x = π/3 par y = 0 se C = −2, aur y = cosx − 2cos²x.

Q10 Find the general solution of dy by dx plus y cotangent x equal cosine squared x.

Answer (English): The integrating factor is sine x, as in the earlier linear example. Multiplying gives the derivative of y sine x equal to sine x cosine squared x, whose integral is minus cosine cubed x over three. Therefore y sine x equal minus cosine cubed x over three plus C, and the general solution is y equal minus cosine cubed x over three sine x plus C cosecant x. There is no initial condition, so the constant remains in the answer.

Hinglish: IF = sinx, aur (y sinx)' = sinx cos²x. Integrate karke y sinx = −cos³x/3 + C, isliye general solution y = −cos³x/(3sinx) + C cosecx hai.

The quiz on this page covers order and degree, the separable method, the homogeneous substitution and the linear integrating factor. Attempt it after writing the ten answers above.