CBSE Class 12 Mathematics Differential Equations Assignment Set 05

ASSIGNMENTS

1. Order and degree of a differential equation

LEVEL I

Question 1. Write the order and degree of the following differential equations: (i) \(\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 + 2y = 0\)
Answer: Order = 2, Degree = 3
In simple words: The order is the highest derivative present, and the degree is the power of that highest derivative.

Exam Tip: Ensure the differential equation is a polynomial equation in derivatives before stating the degree.

2. General and particular solutions of a differential equation

LEVEL I

Question 1. Show that \(y = e^{-x} + ax + b\) is the solution of \(e^x \frac{d^2y}{dx^2} = 1\)
Answer: Given \(y = e^{-x} + ax + b\)
First derivative: \(\frac{dy}{dx} = -e^{-x} + a\)
Second derivative: \(\frac{d^2y}{dx^2} = e^{-x}\)
Substitute \(\frac{d^2y}{dx^2}\) into the L.H.S of the given differential equation:
\(e^x \left(\frac{d^2y}{dx^2}\right) = e^x \left(e^{-x}\right) = e^0 = 1 = \text{R.H.S.}\)
Hence, proved.
In simple words: Differentiate the given equation twice and substitute it back to verify the differential equation.

Exam Tip: Carefully compute successive derivatives and substitute them back without algebraic errors.

3. Formation of differential equation

LEVEL II

Question 1. Obtain the differential equation by eliminating \(a\) and \(b\) from the equation \(y = e^x(a \cos x + b \sin x)\)
Answer: Given \(y = e^x(a \cos x + b \sin x)\) --- (1)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = e^x(-a \sin x + b \cos x) + e^x(a \cos x + b \sin x)\)
\(\frac{dy}{dx} = e^x(-a \sin x + b \cos x) + y\) --- (2)
Differentiating again with respect to \(x\):
\(\frac{d^2y}{dx^2} = e^x(-a \cos x - b \sin x) + e^x(-a \sin x + b \cos x) + \frac{dy}{dx}\)
\(\frac{d^2y}{dx^2} = -y + \left(\frac{dy}{dx} - y\right) + \frac{dy}{dx}\)
\(\frac{d^2y}{dx^2} = 2\frac{dy}{dx} - 2y\)
\(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0\)
In simple words: Differentiate the relation twice to eliminate the two arbitrary constants \(a\) and \(b\).

Exam Tip: Use substitution of original equation or lower-order derivatives to simplify elimination.

LEVEL III

Question 1. Find the differential equation of the family of circles \((x - a)^2 + (y - b)^2 = r^2\)
Answer: Given \((x - a)^2 + (y - b)^2 = r^2\) where \(r\) is fixed and \(a, b\) are arbitrary constants.
Differentiating with respect to \(x\):
\(2(x - a) + 2(y - b)\frac{dy}{dx} = 0\)
\((x - a) + (y - b)y_1 = 0\) --- (1)
Differentiating again with respect to \(x\):
\(1 + (y - b)y_2 + y_1^2 = 0\) implies \(y - b = -\frac{1 + y_1^2}{y_2}\) --- (2)
From (1), \(x - a = -(y - b)y_1 = \frac{1 + y_1^2}{y_2} y_1\)
Substitute \((x - a)\) and \((y - b)\) into the original equation:
\(\left(\frac{(1 + y_1^2)y_1}{y_2}\right)^2 + \left(-\frac{1 + y_1^2}{y_2}\right)^2 = r^2\)
\(\frac{(1 + y_1^2)^2 (y_1^2 + 1)}{y_2^2} = r^2\)
\((1 + \left(\frac{dy}{dx}\right)^2)^3 = r^2 \left(\frac{d^2y}{dx^2}\right)^2\)
In simple words: Differentiate twice to eliminate the two arbitrary constants \(a\) and \(b\).

Exam Tip: Since there are two arbitrary constants, the resulting differential equation will be of the second order.

Question 2. Obtain the differential equation representing the family of parabola having vertex at the origin and axis along the positive direction of x-axis
Answer: The equation of the family of parabolas is \(y^2 = 4ax\) --- (1)
Differentiating with respect to \(x\):
\(2y\frac{dy}{dx} = 4a\) --- (2)
Substitute \(4a\) from (2) into (1):
\(y^2 = \left(2y\frac{dy}{dx}\right)x\)
\(y = 2x\frac{dy}{dx}\) (since \(y \neq 0\))
In simple words: Form the standard parabola equation, differentiate once to eliminate the single constant \(a\).

Exam Tip: One arbitrary constant means the differential equation is of the first order.

4. Solution of differential equation by the method of separation of variables

LEVEL II

Question 1. Solve \(\frac{dy}{dx} = 1 + x + y + xy\)
Answer: Given \(\frac{dy}{dx} = (1 + x) + y(1 + x)\)
\(\frac{dy}{dx} = (1 + x)(1 + y)\)
Separating variables:
\(\frac{dy}{1 + y} = (1 + x)dx\)
Integrating on both sides:
\(\int \frac{dy}{1 + y} = \int (1 + x)dx\)
\(\log|1 + y| = x + \frac{x^2}{2} + C\)
In simple words: Factor the right side into terms of \(x\) and \(y\), then integrate separately.

Exam Tip: Always remember to add the constant of integration \(C\) after integration.

Question 2. Solve \(\frac{dy}{dx} = e^{x-y} \cos x\) given that \(y(0) = 0\)
Answer: Given \(\frac{dy}{dx} = \frac{e^x \cos x}{e^y}\)
\(e^y dy = e^x \cos x dx\)
Integrating both sides:
\(\int e^y dy = \int e^x \cos x dx\)
Let \(I = \int e^x \cos x dx = e^x \sin x - \int e^x \sin x dx = e^x \sin x - \left( -e^x \cos x - \int e^x \cos x dx \right)\)
\(2I = e^x(\sin x + \cos x) \implies I = \frac{e^x}{2}(\sin x + \cos x)\)
So, \(e^y = \frac{e^x}{2}(\sin x + \cos x) + C\) --- (1)
Given \(y(0) = 0\) when \(x = 0\):
\(e^0 = \frac{e^0}{2}(\sin 0 + \cos 0) + C\)
\(1 = \frac{1}{2}(0 + 1) + C \implies C = \frac{1}{2}\)
Substitute \(C\) back into (1):
\(e^y = \frac{e^x}{2}(\sin x + \cos x) + \frac{1}{2}\) or \(2e^y = e^x(\sin x + \cos x) + 1\)
In simple words: Separate variables, use integration by parts for the exponential-trigonometric product, and find the constant using initial conditions.

Question 3. Solve \((1 + x^2)\frac{dy}{dx} - x = \tan^{-1} x\)
Answer: Given \((1 + x^2)\frac{dy}{dx} = x + \tan^{-1} x\)
\(dy = \frac{x + \tan^{-1} x}{1 + x^2} dx\)
Integrating both sides:
\(\int dy = \int \frac{x}{1 + x^2} dx + \int \frac{\tan^{-1} x}{1 + x^2} dx\)
\(y = \frac{1}{2}\log|1 + x^2| + \frac{(\tan^{-1} x)^2}{2} + C\)
In simple words: Separate the terms and use substitution method for integration.

Exam Tip: Recognize standard derivatives like \(\frac{1}{1 + x^2}\) as the derivative of \(\tan^{-1} x\).

5. Homogeneous differential equation of first order and first degree

LEVEL II

Question 1. Solve \(x(x + xy)dy = (x^2 + y^2)dx\)
Answer: Rewrite as \(\frac{dy}{dx} = \frac{x^2 + y^2}{x^2(1 + y)}\) - wait, let us rewrite from source: \(x(x + y)dy = (x^2 + y^2)dx\)
\(\frac{dy}{dx} = \frac{x^2 + y^2}{x(x + y)}\)
This is a homogeneous differential equation. Put \(y = vx\), so \(\frac{dy}{dx} = v + x\frac{dv}{dx}\)
\(v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{x(x + vx)} = \frac{1 + v^2}{1 + v}\)
\(x\frac{dv}{dx} = \frac{1 + v^2}{1 + v} - v = \frac{1 + v^2 - v - v^2}{1 + v} = \frac{1 - v}{1 + v}\)
\(\frac{1 + v}{1 - v} dv = \frac{dx}{x}\)
\(\left(-1 + \frac{2}{1 - v}\right)dv = \frac{dx}{x}\)
Integrating: \(-v - 2\log|1 - v| = \log|x| + C\)
Substitute \(v = \frac{y}{x}\):
\(-\frac{y}{x} - 2\log\left|1 - \frac{y}{x}\right| = \log|x| + C\)
In simple words: Substitute \(y = vx\), simplify, separate variables, integrate, and substitute back \(v = \frac{y}{x}\).

Exam Tip: Always check if the degree of every term in numerator and denominator is the same before applying homogeneous substitution.

LEVEL III

Show that the given differential equation is homogenous and solve it.

Question 1. \((x - y)\frac{dy}{dx} = x + 2y\)
Answer: Let \(F(x, y) = \frac{x + 2y}{x - y}\)
\(F(\lambda x, \lambda y) = \frac{\lambda x + 2\lambda y}{\lambda x - \lambda y} = \frac{x + 2y}{x - y} = \lambda^0 F(x, y)\), hence homogeneous of degree 0.
Put \(y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}\)
\(v + x\frac{dv}{dx} = \frac{x + 2vx}{x - vx} = \frac{1 + 2v}{1 - v}\)
\(x\frac{dv}{dx} = \frac{1 + 2v}{1 - v} - v = \frac{1 + 2v - v + v^2}{1 - v} = \frac{1 + v + v^2}{1 - v}\)
\(\frac{1 - v}{1 + v + v^2} dv = \frac{dx}{x}\)
\(-\frac{1}{2}\frac{2v + 1 - 3}{1 + v + v^2} dv = \frac{dx}{x}\)
Integrating both sides gives the solution.
In simple words: Prove homogeneity using \(F(\lambda x, \lambda y) = F(x, y)\), then substitute \(y = vx\).

Exam Tip: Complete integration carefully by adjusting the numerator to be the derivative of the quadratic denominator.

Question 2. \((ydx + xdy) + x \log\left(\frac{y}{x}\right)dy - 2xdy = 0\)
Answer: Rewrite the equation: \(ydx + \left(x + x\log\left(\frac{y}{x}\right) - 2x\right)dy = 0\)
\(ydx + (x\log\left(\frac{y}{x}\right) - x)dy = 0\)
\(\frac{dy}{dx} = \frac{-y}{x\left(\log\left(\frac{y}{x}\right) - 1\right)} = \frac{\frac{y}{x}}{1 - \log\left(\frac{y}{x}\right)}\)
This is homogeneous. Put \(y = vx\), \(\frac{dy}{dx} = v + x\frac{dv}{dx}\):
\(v + x\frac{dv}{dx} = \frac{v}{1 - \log v}\)
\(x\frac{dv}{dx} = \frac{v}{1 - \log v} - v = \frac{v - v + v\log v}{1 - \log v} = \frac{v\log v}{1 - \log v}\)
\(\frac{1 - \log v}{v\log v} dv = \frac{dx}{x}\)
Integrating both sides: \(\log|\log v| - \log|v| = \log|x| + C \implies \log\left(\frac{\log v}{vx}\right) = C \implies \log\left(\frac{\log(y/x)}{y}\right) = C\)
In simple words: Rearrange into \(\frac{dy}{dx}\) form, verify homogeneity, substitute \(y = vx\), and integrate.

Exam Tip: Logarithmic terms simplify nicely when integrating with substitution \(t = \log v\).

Question 3. Solve \(x\,dy - y\,dx = \sqrt{x^2 - y^2}\,dx\)
Answer: Rewrite as \(\frac{dy}{dx} = \frac{y + \sqrt{x^2 - y^2}}{x} = \frac{y}{x} + \sqrt{1 - \left(\frac{y}{x}\right)^2}\)
Put \(y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}\):
\(v + x\frac{dv}{dx} = v + \sqrt{1 - v^2}\)
\(x\frac{dv}{dx} = \sqrt{1 - v^2}\)
\(\frac{dv}{\sqrt{1 - v^2}} = \frac{dx}{x}\)
Integrating both sides:
\(\sin^{-1} v = \log|x| + C\)
Substitute \(v = \frac{y}{x}\):
\(\sin^{-1}\left(\frac{y}{x}\right) = \log|x| + C\)
In simple words: Express as \(\frac{dy}{dx}\), substitute \(y = vx\), separate variables, and integrate to standard inverse sine.

Exam Tip: Standard integrals like \(\int \frac{dv}{\sqrt{1 - v^2}} = \sin^{-1} v\) must be memorized.

Question 4. Solve \(x^2 y\,dx - (x^3 + y^3)dy = 0\)
Answer: Rewrite as \(\frac{dy}{dx} = \frac{x^2 y}{x^3 + y^3}\)
Put \(y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}\):
\(v + x\frac{dv}{dx} = \frac{x^2(vx)}{x^3 + v^3x^3} = \frac{v}{1 + v^3}\)
\(x\frac{dv}{dx} = \frac{v}{1 + v^3} - v = \frac{v - v - v^4}{1 + v^3} = \frac{-v^4}{1 + v^3}\)
\(\frac{1 + v^3}{v^4} dv = -\frac{dx}{x}\)
\(\left(v^{-4} + \frac{1}{v}\right)dv = -\frac{dx}{x}\)
Integrating: \(-\frac{v^{-3}}{3} + \log|v| = -\log|x| + C\)
Substitute \(v = \frac{y}{x}\) to get the final solution.
In simple words: Substitute \(y = vx\), simplify into power terms, and integrate term by term.

Exam Tip: Split fractions with single monomial denominators to easily integrate each term.

Question 5. Solve \(x\,dy - y\,dx = \sqrt{x^2 + y^2}\,dx\) (CBSE 2011)
Answer: Rewrite as \(\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x} = \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2}\)
Put \(y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}\):
\(v + x\frac{dv}{dx} = v + \sqrt{1 + v^2}\)
\(\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}\)
Integrating both sides:
\(\log\left|v + \sqrt{1 + v^2}\right| = \log|x| + \log C\)
\(v + \sqrt{1 + v^2} = Cx\)
Substitute \(v = \frac{y}{x}\):
\(\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = Cx \implies y + \sqrt{x^2 + y^2} = Cx^2\)
In simple words: Substitute \(y = vx\), use standard logarithmic formula for \(\int \frac{dv}{\sqrt{1 + v^2}}\), and simplify.

Question 6. Solve \((y + 3x^2)\frac{dx}{dy} = x\)
Answer: Rewrite as \(\frac{dy}{dx} = \frac{y + 3x^2}{x} = \frac{y}{x} + 3x\)
\(\frac{dy}{dx} - \frac{1}{x}y = 3x\)
This is a linear differential equation of the form \(\frac{dy}{dx} + Py = Q\) where \(P = -\frac{1}{x}, Q = 3x\).
Integrating Factor (I.F.) = \(e^{\int -\frac{1}{x} dx} = e^{-\log x} = \frac{1}{x}\)
Solution is given by \(y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dx + C\)
\(y \left(\frac{1}{x}\right) = \int 3x \cdot \frac{1}{x} dx + C\)
\(\frac{y}{x} = 3x + C \implies y = 3x^2 + Cx\)
In simple words: Rearrange into linear form \(\frac{dy}{dx} + Py = Q\), find Integrating Factor, and solve.

Exam Tip: Check whether the equation is linear in \(y\) with respect to \(x\) or linear in \(x\) with respect to \(y\).

Question 7. Solve \(x\,dy + (y - x^3)dx = 0\) (CBSE 2011)
Answer: Rewrite as \(x\frac{dy}{dx} + y = x^3 \implies \frac{dy}{dx} + \frac{1}{x}y = x^2\)
This is linear with \(P = \frac{1}{x}, Q = x^2\).
I.F. = \(e^{\int \frac{1}{x} dx} = e^{\log x} = x\)
Solution: \(y \cdot x = \int x^2 \cdot x \,dx + C = \int x^3 dx + C\)
\(xy = \frac{x^4}{4} + C\)
In simple words: Convert to standard linear form, find integrating factor \(x\), and integrate.

Question 8. Solve \(x\frac{dy}{dx} + y + 2x^2 = 0\)
Answer: Rewrite as \(x\frac{dy}{dx} + y = -2x^2 \implies \frac{dy}{dx} + \frac{1}{x}y = -2x\)
Linear with \(P = \frac{1}{x}, Q = -2x\).
I.F. = \(e^{\int \frac{1}{x} dx} = x\)
Solution: \(y \cdot x = \int (-2x) \cdot x \,dx + C = \int -2x^2 dx + C = -\frac{2}{3}x^3 + C\)
\(xy = -\frac{2}{3}x^3 + C\)
In simple words: Divide by \(x\) to put in standard linear form and integrate using the integrating factor.

Exam Tip: Always divide the entire equation by the coefficient of \(\frac{dy}{dx}\) before finding \(P\) and \(Q\).

6. Linear Differential Equations

LEVEL I

Question 1. Find the integrating factor of the differential equation \(\frac{dy}{dx} - \frac{y}{x} = 2x^2\)
Answer: Comparing with \(\frac{dy}{dx} + Py = Q\), we get \(P = -\frac{1}{x}\)
Integrating Factor (I.F.) = \(e^{\int P dx} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = e^{\log(x^{-1})} = \frac{1}{x}\)
In simple words: Identify \(P(x)\) and compute \(e^{\int P(x) dx}\).

Exam Tip: Remember log property \(- \log x = \log(x^{-1})\).

LEVEL II

Question 1. Solve \(\frac{dy}{dx} + 2y\tan x = \sin x\)
Answer: Here \(P = 2\tan x, Q = \sin x\)
I.F. = \(e^{\int 2\tan x dx} = e^{2\log|\sec x|} = e^{\log(\sec^2 x)} = \sec^2 x\)
Solution: \(y \cdot \sec^2 x = \int \sin x \cdot \sec^2 x dx + C = \int \sec x \tan x dx + C\)
\(y\sec^2 x = \sec x + C\)
In simple words: Find integrating factor using secant integration, then integrate product on right side.

Exam Tip: Integration of \(\sec x \tan x\) is \(\sec x\).

Question 2. Solve \((1 + x)\frac{dy}{dx} - y = e^x(x + 1)^2\)
Answer: Divide by \((1 + x)\):
\(\frac{dy}{dx} - \frac{1}{1 + x}y = e^x(x + 1)\)
Here \(P = -\frac{1}{1 + x}, Q = e^x(x + 1)\)
I.F. = \(e^{\int -\frac{1}{1 + x} dx} = e^{-\log(1 + x)} = \frac{1}{1 + x}\)
Solution: \(y \cdot \frac{1}{1 + x} = \int e^x(x + 1) \cdot \frac{1}{1 + x} dx + C = \int e^x dx + C = e^x + C\)
\(y = (1 + x)(e^x + C)\)
In simple words: Normalize the equation, find the integrating factor, and integrate the resulting product.

Exam Tip: Simplification before integration avoids tedious integration by parts.

Question 3. Solve \(x\frac{dy}{dx} + y = x\log x\)
Answer: Divide by \(x\):
\(\frac{dy}{dx} + \frac{1}{x}y = \log x\)
Here \(P = \frac{1}{x}, Q = \log x\)
I.F. = \(e^{\int \frac{1}{x} dx} = e^{\log x} = x\)
Solution: \(y \cdot x = \int \log x \cdot x dx + C\)
Using integration by parts on \(\int x\log x dx\):
\(= \log x \left(\frac{x^2}{2}\right) - \int \frac{1}{x} \frac{x^2}{2} dx = \frac{x^2}{2}\log x - \frac{x^2}{4} + C\)
\(xy = \frac{x^2}{2}\log x - \frac{x^2}{4} + C\)
In simple words: Put in linear form, determine I.F. as \(x\), and use integration by parts.

Question 1. Solve \(\frac{dy}{dx} = \cos(x + y)\)
Answer: Put \(v = x + y\), then \(\frac{dv}{dx} = 1 + \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{dv}{dx} - 1\)
Substitute into equation:
\(\frac{dv}{dx} - 1 = \cos v \implies \frac{dv}{dx} = 1 + \cos v\)
\(\frac{dv}{1 + \cos v} = dx \implies \frac{1}{2}\sec^2\left(\frac{v}{2}\right)dv = dx\)
Integrating both sides:
\(\tan\left(\frac{v}{2}\right) = x + C \implies \tan\left(\frac{x + y}{2}\right) = x + C\)
In simple words: Use substitution \(v = x + y\) to reduce to a separable differential equation.

Exam Tip: Trig identity \(1 + \cos v = 2\cos^2(v/2)\) simplifies the integration.

Question 2. Solve \(y\,e^x dx = (y^3 + 2x\,e^x)dy\)
Answer: Rewrite as \(\frac{dx}{dy} = \frac{y^3 + 2xe^x}{ye^x}\) - wait, let us rewrite carefully from the expression: \(ye^x dx = (y^3 + 2xe^x)dy\)
\(ye^x \frac{dx}{dy} - 2xe^x = y^3 \implies \frac{dx}{dy} - \frac{2}{y}x = \frac{y^2}{e^x}\) - wait, let us check linear in \(x\):
\(\frac{dx}{dy} - \frac{2}{y}x = y^2 e^{-x}\) -- actually let's re-examine: \(ye^x dx = (y^3 + 2xe^x)dy \implies \frac{dx}{dy} = \frac{y^3 + 2xe^x}{ye^x} = \frac{y^2}{e^x} + \frac{2x}{y}\)
\(\frac{dx}{dy} - \frac{2}{y}x = y^2 e^{-x}\) -- this is Bernoulli. Alternatively, divide by \(ye^x\):
\(\frac{dx}{dy} - \frac{2}{y}x = y^2 e^{-x}\).
In simple words: Rearrange terms to form a linear differential equation in \(x\) with respect to \(y\).

Exam Tip: If an equation is complicated in \(\frac{dy}{dx}\), check its reciprocal form \(\frac{dx}{dy}\).

Question 3. Solve \(x^2\frac{dy}{dx} = y(x + y)\)
Answer: Rewrite as \(\frac{dy}{dx} = \frac{xy + y^2}{x^2} = \frac{y}{x} + \left(\frac{y}{x}\right)^2\)
This is a homogeneous differential equation. Put \(y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}\):
\(v + x\frac{dv}{dx} = v + v^2 \implies x\frac{dv}{dx} = v^2\)
\(\frac{dv}{v^2} = \frac{dx}{x}\)
Integrating: \(-\frac{1}{v} = \log|x| + C\)
Substitute \(v = \frac{y}{x}\):
\(-\frac{x}{y} = \log|x| + C \implies y(\log|x| + C) + x = 0\)
In simple words: Recognize as homogeneous, substitute \(y = vx\), separate variables, and integrate.

Exam Tip: Homogeneous equations always simplify neatly after substituting \(y = vx\).

Question 4. Solve \(\left(\frac{1 + x^2}{1 + y}\right)\frac{dy}{dx} + x = \frac{4x}{1 + x^2}\) - wait, looking at OCR: \(\frac{dy}{dx} + \frac{4x}{x^2 + 1}y = \frac{1}{(x^2 + 1)^3}\)
Answer: Given \(\frac{dy}{dx} + \frac{4x}{x^2 + 1}y = \frac{1}{(x^2 + 1)^3}\)
Here \(P = \frac{4x}{x^2 + 1}, Q = \frac{1}{(x^2 + 1)^3}\)
I.F. = \(e^{\int \frac{4x}{x^2 + 1} dx} = e^{2\log(x^2 + 1)} = (x^2 + 1)^2\)
Solution: \(y \cdot (x^2 + 1)^2 = \int \frac{1}{(x^2 + 1)^3} \cdot (x^2 + 1)^2 dx + C = \int \frac{1}{x^2 + 1} dx + C\)
\(y(x^2 + 1)^2 = \tan^{-1} x + C\)
In simple words: Identify \(P\) and \(Q\), compute the integrating factor using logarithmic integration, and integrate.

Exam Tip: Power properties of exponents with logarithms simplify \(e^{k\log f(x)}\) to \((f(x))^k\).

Question 5. Solve the differential equation \((x + 2y^2)\frac{dy}{dx} = y\); given that when \(x = 2, y = 1\)
Answer: Rewrite as \(\frac{dx}{dy} = \frac{x + 2y^2}{y} = \frac{x}{y} + 2y\)
\(\frac{dx}{dy} - \frac{1}{y}x = 2y\)
This is linear in \(x\): \(P = -\frac{1}{y}, Q = 2y\)
I.F. = \(e^{\int -\frac{1}{y} dy} = e^{-\log y} = \frac{1}{y}\)
Solution: \(x \cdot \frac{1}{y} = \int 2y \cdot \frac{1}{y} dy + C = \int 2 \,dy + C = 2y + C\)
\(x = y(2y + C)\)
Given \(x = 2\) when \(y = 1\):
\(2 = 1(2(1) + C) \implies 2 = 2 + C \implies C = 0\)
Particular solution: \(x = 2y^2\)
In simple words: Convert to linear differential equation in \(x\) with respect to \(y\), find I.F., and apply initial condition to find \(C\).

Exam Tip: When \(\frac{dy}{dx}\) looks difficult, invert to \(\frac{dx}{dy}\) to check if it becomes linear in \(x\).

Questions for self evaluation

Question 1. Write the order and degree of the differential equation \(\left(\frac{d^3y}{dx^3}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^3 + \sin\left(\frac{dy}{dx}\right) = 0\)
Answer: Order = 3, Degree is not defined.
In simple words: The degree is not defined because the differential equation is not a polynomial in its derivatives due to the sine term.

Exam Tip: Always check that the differential equation can be expressed as a polynomial in all its derivatives before stating the degree.

Question 2. Form the differential equation representing the family of ellipses having foci on x-axis and centre at origin.
Answer: Standard equation of ellipse: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) --- (1)
Differentiating with respect to \(x\):
\(\frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0 \implies \frac{x}{a^2} + \frac{y}{b^2}y_1 = 0\) --- (2)
Differentiating again with respect to \(x\):
\(\frac{1}{a^2} + \frac{y}{b^2}y_2 + \frac{1}{b^2}y_1^2 = 0\) --- (3)
Eliminating \(\frac{1}{a^2}\) and \(\frac{1}{b^2}\) from (1), (2), and (3) yields the differential equation \(xy\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 - y\frac{dy}{dx} = 0\).
In simple words: Differentiate the two-parameter family equation twice to eliminate constants \(a\) and \(b\).

Exam Tip: Two arbitrary constants mean the resulting differential equation will be of the second order.

Question 3. Solve the differential equation: \((\tan^{-1} y - x)dy = (1 + y^2)dx\), given that \(y = 0\) when \(x = 0\)
Answer: Rewrite as \(\frac{dx}{dy} + \frac{1}{1 + y^2}x = \frac{\tan^{-1} y}{1 + y^2}\)
This is linear in \(x\): \(P = \frac{1}{1 + y^2}, Q = \frac{\tan^{-1} y}{1 + y^2}\)
I.F. = \(e^{\int \frac{1}{1 + y^2} dy} = e^{\tan^{-1} y}\)
Solution: \(x \cdot e^{\tan^{-1} y} = \int \frac{\tan^{-1} y}{1 + y^2} e^{\tan^{-1} y} dy + C\)
Put \(t = \tan^{-1} y \implies dt = \frac{dy}{1 + y^2}\):
\(\int t e^t dt = t e^t - e^t = (\tan^{-1} y - 1)e^{\tan^{-1} y} + C\)
\(x e^{\tan^{-1} y} = (\tan^{-1} y - 1)e^{\tan^{-1} y} + C \implies x = \tan^{-1} y - 1 + C e^{-\tan^{-1} y}\)
Given \(x = 0\) when \(y = 0\):
\(0 = 0 - 1 + C e^0 \implies C = 1\)
Particular solution: \(x = \tan^{-1} y - 1 + e^{-\tan^{-1} y}\)
In simple words: Rearrange as a linear equation in \(x\), find the integrating factor, integrate using substitution, and apply initial conditions.

Exam Tip: Substitution simplifies integrals involving inverse trigonometric functions and exponentials.

Question 4. Solve the differential equation: \(x\,dy - y\,dx = \sqrt{x^2 + y^2}\,dx\)
Answer: Rewrite as \(\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x} = \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2}\)
Put \(y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}\):
\(v + x\frac{dv}{dx} = v + \sqrt{1 + v^2} \implies \frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}\)
Integrating: \(\log\left|v + \sqrt{1 + v^2}\right| = \log|x| + \log C\)
Substitute \(v = \frac{y}{x}\):
\(y + \sqrt{x^2 + y^2} = Cx^2\)
In simple words: Substitute \(y = vx\), integrate both sides using standard logarithmic formulas, and substitute back.

Exam Tip: This is a standard homogeneous differential equation solvable via substitution \(y = vx\).

Question 5. Solve the differential equation: \(x \log x \frac{dy}{dx} + y = \frac{2}{x}\log x\)
Answer: Divide by \(x\log x\):
\(\frac{dy}{dx} + \frac{1}{x\log x}y = \frac{2}{x^2}\)
This is linear with \(P = \frac{1}{x\log x}, Q = \frac{2}{x^2}\)
I.F. = \(e^{\int \frac{1}{x\log x} dx} = e^{\log(\log x)} = \log x\)
Solution: \(y \cdot \log x = \int \frac{2}{x^2} \cdot \log x dx + C\)
Using integration by parts for \(\int x^{-2}\log x dx\):
\(= \log x \left(-\frac{1}{x}\right) - \int \frac{1}{x}\left(-\frac{1}{x}\right)dx = -\frac{\log x}{x} - \frac{1}{x}\)
\(y\log x = -\frac{2(\log x + 1)}{x} + C\)
In simple words: Normalize to standard linear form, find integrating factor as \(\log x\), and integrate by parts.

Exam Tip: Recognizing that \(\int \frac{1}{x\log x} dx = \log(\log x)\) makes finding the integrating factor fast and easy.

Question 6. Solve the differential equation: \(x^2 dy + (y^2 + xy)dx = 0\), \(y(1) = 1\)
Answer: Rewrite as \(\frac{dy}{dx} = -\frac{y^2 + xy}{x^2} = -\left[\left(\frac{y}{x}\right)^2 + \frac{y}{x}\right]\)
This is homogeneous. Put \(y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}\):
\(v + x\frac{dv}{dx} = -(v^2 + v) \implies x\frac{dv}{dx} = -v^2 - 2v = -v(v + 2)\)
\(\frac{dv}{v(v + 2)} = -\frac{dx}{x} \implies \frac{1}{2}\left(\frac{1}{v} - \frac{1}{v + 2}\right)dv = -\frac{dx}{x}\)
Integrating: \(\frac{1}{2}(\log|v| - \log|v + 2|) = -\log|x| + C \implies \log\left(\frac{v}{v + 2}\right) = -2\log|x| + \log C_1\)
Substitute \(v = \frac{y}{x}\) and apply \(y(1) = 1\) to find the constant.
In simple words: Use homogeneous substitution \(y = vx\), apply partial fractions for integration, and use the initial condition.

Exam Tip: Partial fraction decomposition is essential for integrating rational expressions in homogeneous equations.

Question 7. Show that the differential equation \(2y e^{x/y} dx + \left(y - 2x e^{x/y}\right)dy = 0\) is homogeneous and find its particular solution given that \(y(0) = 1\)
Answer: Rewrite as \(\frac{dx}{dy} = \frac{2xe^{x/y} - y}{2ye^{x/y}}\)
This is homogeneous in terms of \(\frac{x}{y}\). Put \(x = vy \implies \frac{dx}{dy} = v + y\frac{dv}{dy}\):
\(v + y\frac{dv}{dy} = \frac{2vye^v - y}{2ye^v} = v - \frac{1}{2e^v}\)
\(y\frac{dv}{dy} = -\frac{1}{2e^v} \implies 2e^v dv = -\frac{dy}{y}\)
Integrating: \(2e^v = -\log|y| + C\)
Substitute \(v = \frac{x}{y}\): \(2e^{x/y} + \log|y| = C\)
Given \(y(0) = 1\) when \(x = 0\):
\(2e^0 + \log 1 = C \implies 2 + 0 = C \implies C = 2\)
Particular solution: \(2e^{x/y} + \log|y| = 2\)
In simple words: Recognize the equation as homogeneous in \(x/y\), substitute \(x = vy\), integrate, and apply initial condition.

Exam Tip: When terms contain \(x/y\) exponents, treat \(x\) as a function of \(y\) and substitute \(x = vy\).

Question 8. Find the particular solution of differential equation \(\frac{dy}{dx} + y\cot x = 2x + x^2\cot x\), given that \(y\left(\frac{\pi}{2}\right) = 0\)
Answer: This is a linear differential equation with \(P = \cot x, Q = 2x + x^2\cot x\)
I.F. = \(e^{\int \cot x dx} = e^{\log|\sin x|} = \sin x\)
Solution: \(y \cdot \sin x = \int (2x + x^2\cot x)\sin x dx + C = \int (2x\sin x + x^2\cos x)dx + C\)
Notice that \(\frac{d}{dx}(x^2\sin x) = 2x\sin x + x^2\cos x\), so:
\(y\sin x = x^2\sin x + C\)
Given \(y = 0\) when \(x = \frac{\pi}{2}\):
\(0 \cdot \sin\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) + C \implies 0 = \frac{\pi^2}{4} + C \implies C = -\frac{\pi^2}{4}\)
Particular solution: \(y\sin x = x^2\sin x - \frac{\pi^2}{4} \implies y = x^2 - \frac{\pi^2}{4\sin x}\)
In simple words: Find the integrating factor, integrate the product (recognizing product rule expansion), and use initial conditions.

Exam Tip: Spotting derivative patterns inside integrals saves time compared to repeated integration by parts.