CBSE Class 12 Mathematics Differential Equations Assignment Set 04

1. Order and degree of a differential equation

LEVEL I

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

Exam Tip: Make sure the differential equation is a polynomial equation in derivatives before determining its 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 equation: \(y = e^{-x} + ax + b\)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = -e^{-x} + a\)
Differentiating again with respect to \(x\):
\(\frac{d^2y}{dx^2} = e^{-x}\)
Substituting this into the L.H.S. of the given differential equation:
\(e^x \frac{d^2y}{dx^2} = e^x \cdot e^{-x} = e^0 = 1 = \text{R.H.S.}\)
Hence, proved.
In simple words: Substitute the given function and its derivatives into the differential equation to check if both sides are equal.

Exam Tip: Successive differentiation must be performed carefully, paying close attention to signs of exponential functions.

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 equation: \(y = e^x(a \cos x + b \sin x)\) -- (1)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = e^x(a \cos x + b \sin x) + e^x(-a \sin x + b \cos x)\)
\(\frac{dy}{dx} = y + e^x(-a \sin x + b \cos x)\)
\(\frac{dy}{dx} - y = e^x(-a \sin x + b \cos x)\) -- (2)
Differentiating again with respect to \(x\):
\(\frac{d^2y}{dx^2} - \frac{dy}{dx} = e^x(-a \sin x + b \cos x) + e^x(-a \cos x - b \sin x)\)
\(\frac{d^2y}{dx^2} - \frac{dy}{dx} = \left(\frac{dy}{dx} - y\right) - y\) (using equations (1) and (2))
\(\frac{d^2y}{dx^2} - \frac{dy}{dx} = \frac{dy}{dx} - 2y\)
\(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0\)
In simple words: Differentiate the equation as many times as there are arbitrary constants, then eliminate those constants.

Exam Tip: Reuse the original equation or its first derivative during simplification to quickly eliminate arbitrary constants.

LEVEL III

Question 1. Find the differential equation of the family of circles \((x - a)^2 + (y - b)^2 = r^2\).
Answer: Given equation: \((x - a)^2 + (y - b)^2 = r^2\) where \(r\) is fixed and \(a, b\) are arbitrary constants.
Differentiating with respect to \(x\):
\(2(x - a) + 2(y - b)\frac{dy}{dx} = 0\)
\((x - a) + (y - b)y' = 0\) -- (1)
Differentiating again with respect to \(x\):
\(1 + (y - b)y'' + y'(y') = 0\)
\((y - b)y'' + 1 + (y')^2 = 0 \implies y - b = -\frac{1 + (y')^2}{y''}\) -- (2)
From (1), we have \(x - a = -(y - b)y'\). Substituting (2) into this:
\(x - a = \frac{1 + (y')^2}{y''} y'\)
Substituting \((x - a)\) and \((y - b)\) back into the original circle equation:
\(\left(\frac{(1 + (y')^2)y'}{y''}\right)^2 + \left(-\frac{1 + (y')^2}{y''}\right)^2 = r^2\)
\(\frac{(1 + (y')^2)^2 [ (y')^2 + 1 ]}{(y'')^2} = r^2\)
\(\left(1 + \left(\frac{dy}{dx}\right)^2\right)^3 = r^2 \left(\frac{d^2y}{dx^2}\right)^2\)
In simple words: Eliminate the two center parameters \(a\) and \(b\) by differentiating twice and substituting back.

Exam Tip: Keep track of the number of arbitrary constants. Here \(a\) and \(b\) are constants to eliminate, whereas \(r\) is a fixed radius parameter.

Question 2. Obtain the differential equation representing the family of parabolas having vertex at the origin and axis along the positive direction of \(x\)-axis.
Answer: The equation of the family of parabolas is \(y^2 = 4ax\) where \(a\) is an arbitrary constant.
Differentiating with respect to \(x\):
\(2y \frac{dy}{dx} = 4a\)
\(y \frac{dy}{dx} = 2a\) -- (1)
Substitute \(2a\) from equation (1) into the original parabola equation:
\(y^2 = \left(y \frac{dy}{dx}\right) 2x\)
\(y = 2x \frac{dy}{dx}\) (since \(y \neq 0\))
In simple words: Find the first derivative, isolate the constant \(a\), and substitute it back to remove it.

Exam Tip: A single arbitrary constant requires a first-order differential equation.

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 equation: \(\frac{dy}{dx} = 1 + x + y(1 + x)\)
\(\frac{dy}{dx} = (1 + x)(1 + y)\)
Separating the variables:
\(\frac{dy}{1 + y} = (1 + x)dx\)
Integrating both sides:
\(\int \frac{dy}{1 + y} = \int (1 + x)dx\)
\(\log|1 + y| = x + \frac{x^2}{2} + C\)
In simple words: Factorize the R.H.S. into terms of \(x\) and \(y\), separate them, and integrate.

Exam Tip: Always add the constant of integration \(C\) immediately after performing integration.

Question 2. Solve \(\frac{dy}{dx} = e^{x-y} \cos x\) given that \(y(0) = 0\).
Answer: Given equation: \(\frac{dy}{dx} = \frac{e^x}{e^y} \cos x\)
Separating the variables:
\(e^y dy = e^x \cos x dx\)
Integrating both sides:
\(\int e^y dy = \int e^x \cos x dx\)
Let \(I = \int e^x \cos x dx = e^x \sin x - \int e^x \sin x dx = e^x \sin x - \left( -e^x \cos x + \int e^x \cos x dx \right)\)
\(2I = e^x(\sin x + \cos x) \implies I = \frac{e^x}{2}(\sin x + \cos x)\)
Thus, \(e^y = \frac{e^x}{2}(\sin x + \cos x) + C\)
Using the condition \(y(0) = 0\) when \(x = 0\):
\(e^0 = \frac{e^0}{2}(\sin 0 + \cos 0) + C \implies 1 = \frac{1}{2}(0 + 1) + C \implies C = \frac{1}{2}\)
Therefore, the particular solution is:
\(e^y = \frac{e^x}{2}(\sin x + \cos x) + \frac{1}{2} \implies 2e^y = e^x(\sin x + \cos x) + 1\)
In simple words: Separate variables, integrate using integration by parts for the exponential-trigonometric product, and find \(C\) using the initial condition.

Question 3. Solve \((1 + x^2)\frac{dy}{dx} - x = \tan^{-1} x\)
Answer: Rearranging the given equation:
\((1 + x^2)\frac{dy}{dx} = x + \tan^{-1} x\)
\(dy = \frac{x + \tan^{-1} x}{1 + x^2} dx\)
Integrating both sides:
\(\int dy = \int \frac{x}{1 + x^2} dx + \int \frac{\tan^{-1} x}{1 + x^2} dx\)
For the first integral, let \(1 + x^2 = t \implies 2x dx = dt\).
For the second integral, let \(\tan^{-1} x = u \implies \frac{1}{1 + x^2} dx = du\).
\(y = \frac{1}{2}\log|1 + x^2| + \frac{(\tan^{-1} x)^2}{2} + C\)
\(2y = \log(1 + x^2) + (\tan^{-1} x)^2 + C_1\)
In simple words: Separate \(dx\) and \(dy\), then use substitution for both terms on the right-hand side.

Exam Tip: Recognize standard derivative forms like \(\frac{1}{1 + x^2}\) to apply substitution smoothly.

5. Homogeneous differential equation of first order and first degree

LEVEL II

Question 1. Solve \((x^2 + xy)dy = (x^2 + y^2)dx\)
Answer: Rewriting as:
\(\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy}\)
This is a homogeneous differential equation. Put \(y = vx \implies \frac{dy}{dx} = v + x \frac{dv}{dx}\).
\(v + x \frac{dv}{dx} = \frac{x^2 + v^2x^2}{x^2 + vx^2} = \frac{1 + v^2}{1 + v}\)
\(x \frac{dv}{dx} = \frac{1 + v^2}{1 + v} - v = \frac{1 + v^2 - v - v^2}{1 + v} = \frac{1 - v}{1 + v}\)
Separating variables:
\(\frac{1 + v}{1 - v} dv = \frac{dx}{x}\)
\(\left(-1 + \frac{2}{1 - v}\right) dv = \frac{dx}{x}\)
Integrating both sides:
\(-v - 2\log|1 - v| = \log|x| + C\)
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 the resulting separable equation in \(v\) and \(x\), integrate, and substitute back \(v = y/x\).

Exam Tip: Splitting algebraic fractions like \(\frac{1+v}{1-v}\) into \(-1 + \frac{2}{1-v}\) makes integration straightforward.

LEVEL III

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

Question 1. \((x - y)\frac{dy}{dx} = x + 2y\)
Answer: Let \(F(x, y) = \frac{x + 2y}{x - y}\).
Replacing \(x\) by \(tx\) and \(y\) by \(ty\):
\(F(tx, ty) = \frac{tx + 2ty}{tx - ty} = \frac{t(x + 2y)}{t(x - y)} = t^0 F(x, y)\)
Thus, the equation is homogeneous of degree zero.
Put \(y = vx \implies \frac{dy}{dx} = v + x \frac{dv}{dx}\).
\(v + x \frac{dv}{dx} = \frac{x + 2vx}{x - vx} = \frac{1 + 2v}{1 - v}\)
\(x \frac{dv}{dx} = \frac{1 + 2v}{1 - v} - v = \frac{1 + 2v - v + v^2}{1 - v} = \frac{1 + v + v^2}{1 - v}\)
Separating variables:
\(\frac{1 - v}{1 + v + v^2} dv = \frac{dx}{x}\)
Note that derivative of denominator is \(2v + 1\). Rewrite numerator: \(1 - v = -\frac{1}{2}(2v + 1) + \frac{3}{2}\).
\(\int \frac{-\frac{1}{2}(2v + 1) + \frac{3}{2}}{v^2 + v + 1} dv = \int \frac{dx}{x}\)
\(-\frac{1}{2}\log(v^2 + v + 1) + \frac{3}{2} \cdot \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \log|x| + C\)
Substitute \(v = \frac{y}{x}\) to get the final solution.
In simple words: Check homogeneity by scaling variables by \(t\), substitute \(y = vx\), and integrate using standard partial fraction / completing the square formulas.

Exam Tip: When integrating rational functions with quadratic denominators where the discriminant is negative, complete the square to use the inverse tangent formula.

Question 2. \((y \log y)dx + (x - log y)dy = 0\)
Answer: Rearranging the given equation:
\((x - \log y)dy = -y \log y dx\)
\(\frac{dx}{dy} = \frac{\log y - x}{y \log y} = \frac{\log y}{y \log y} - \frac{x}{y \log y} = \frac{1}{y} - \frac{x}{y \log y}\)
\(\frac{dx}{dy} + \frac{1}{y \log y} x = \frac{1}{y}\)
This is a linear differential equation of the form \(\frac{dx}{dy} + Px = Q\), where \(P = \frac{1}{y \log y}\) and \(Q = \frac{1}{y}\).
Integrating Factor (\(\text{I.F.}\)) = \(e^{\int \frac{1}{y \log y} dy} = e^{\log(\log y)} = \log y\).
Solution is given by:
\(x \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dy + C\)
\(x \log y = \int \frac{1}{y} \log y dy + C\)
Let \(\log y = t \implies \frac{1}{y}dy = dt\).
\(x \log y = \int t dt + C = \frac{t^2}{2} + C = \frac{(\log y)^2}{2} + C\)
\(2x \log y = (\log y)^2 + C_1\)
In simple words: Treat \(x\) as the dependent variable and \(y\) as the independent variable, convert to linear form, find the integrating factor, and integrate.

Exam Tip: Watch out for equations that are linear in \(x\) (i.e., \(\frac{dx}{dy} + Px = Q\)) rather than linear in \(y\).

Question 3. Solve \(x \, dy - y \, dx = \sqrt{x^2 - y^2} \, dx\)
Answer: Rearranging:
\(x \, dy = (y + \sqrt{x^2 - y^2}) dx\)
\(\frac{dy}{dx} = \frac{y + \sqrt{x^2 - y^2}}{x} = \frac{y}{x} + \sqrt{1 - \left(\frac{y}{x}\right)^2}\)
Put \(y = vx \implies \frac{dy}{dx} = v + x \frac{dv}{dx}\).
\(v + x \frac{dv}{dx} = v + \sqrt{1 - v^2}\)
\(x \frac{dv}{dx} = \sqrt{1 - v^2}\)
Separating variables:
\(\frac{dv}{\sqrt{1 - v^2}} = \frac{dx}{x}\)
Integrating both sides:
\(\sin^{-1}(v) = \log|x| + C\)
Substitute \(v = \frac{y}{x}\):
\(\sin^{-1}\left(\frac{y}{x}\right) = \log|x| + C \implies y = x \sin(\log|x| + C)\)
In simple words: Convert to \(\frac{dy}{dx}\), substitute \(y = vx\), simplify to standard integral form, and substitute back.

Exam Tip: \(\int \frac{1}{\sqrt{1-v^2}} dv = \sin^{-1}(v)\) is a standard integral formula that frequently appears in homogeneous equations.

Question 4. Solve \(x^2 y \, dx - (x^3 + y^3) \, dy = 0\)
Answer: Rewriting as:
\(\frac{dy}{dx} = \frac{x^2 y}{x^3 + y^3}\)
This is a homogeneous differential equation. Put \(y = vx \implies \frac{dy}{dx} = v + x \frac{dv}{dx}\).
\(v + x \frac{dv}{dx} = \frac{x^2(vx)}{x^3 + v^3 x^3} = \frac{v}{1 + v^3}\)
\(x \frac{dv}{dx} = \frac{v}{1 + v^3} - v = \frac{v - v - v^4}{1 + v^3} = \frac{-v^4}{1 + v^3}\)
Separating variables:
\(\frac{1 + v^3}{v^4} dv = -\frac{dx}{x}\)
\(\left(v^{-4} + v^{-1}\right) dv = -\frac{dx}{x}\)
Integrating both sides:
\(-\frac{v^{-3}}{3} + \log|v| = -\log|x| + C\)
\(\log|xy| = \frac{1}{3v^3} + C\)
Substitute \(v = \frac{y}{x}\):
\(\log|xy| = \frac{x^3}{3y^3} + C\)
In simple words: Substitute \(y = vx\), separate variables, integrate term-by-term, and substitute back original variables.

Exam Tip: Splitting \(\frac{1+v^3}{v^4}\) into \(\frac{1}{v^4} + \frac{1}{v}\) makes integration straightforward.

Question 5. Solve \(x \, dy - y \, dx = \sqrt{x^2 + y^2} \, dx\) 
Answer: Rearranging:
\(x \, dy = (y + \sqrt{x^2 + y^2}) dx\)
\(\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x} = \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2}\)
Put \(y = vx \implies \frac{dy}{dx} = v + x \frac{dv}{dx}\).
\(v + x \frac{dv}{dx} = v + \sqrt{1 + v^2}\)
\(x \frac{dv}{dx} = \sqrt{1 + v^2}\)
Separating variables:
\(\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}\)
Integrating both sides:
\(\log\left|v + \sqrt{1 + v^2}\right| = \log|x| + \log C\)
\(v + \sqrt{1 + v^2} = Cx\)
Substitute \(v = \frac{y}{x}\):
\(\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = Cx \implies y + \sqrt{x^2 + y^2} = C x^2\)
In simple words: Use homogeneous substitution \(y = vx\), integrate using standard logarithmic form for square roots, and simplify.

Exam Tip: Remember that \(\int \frac{dv}{\sqrt{1+v^2}} = \log|v + \sqrt{1+v^2}|\).

Question 6. Solve \((y + 3x^2) \frac{dx}{dy} = x\)
Answer: Rewriting 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 in \(y\): \(\frac{dy}{dx} + Py = Q\), where \(P = -\frac{1}{x}\) and \(Q = 3x\).
\(\text{I.F.} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = e^{\log(x^{-1})} = \frac{1}{x}\).
Solution:
\(y \cdot \left(\frac{1}{x}\right) = \int 3x \cdot \left(\frac{1}{x}\right) dx + C\)
\(\frac{y}{x} = \int 3 dx + C = 3x + C\)
\(y = 3x^2 + Cx\)
In simple words: Invert the derivative to make it a standard linear differential equation in \(y\), find the integrating factor, and integrate.

Exam Tip: If an equation looks complex in \(\frac{dx}{dy}\), check if taking \(\frac{dy}{dx}\) transforms it into a standard linear form.

Question 7. Solve \(x \frac{dy}{dx} + y - x^3 = 0\) 
Answer: Rewriting in standard linear form:
\(x \frac{dy}{dx} + y = x^3 \implies \frac{dy}{dx} + \frac{1}{x} y = x^2\)
Here, \(P = \frac{1}{x}\) and \(Q = x^2\).
\(\text{I.F.} = e^{\int \frac{1}{x} dx} = e^{\log x} = x\).
Solution:
\(y \cdot x = \int x^2 \cdot x \, dx + C\)
\(xy = \int x^3 dx + C = \frac{x^4}{4} + C\)
\(y = \frac{x^3}{4} + \frac{C}{x}\)
In simple words: Divide by \(x\) to put the equation in standard linear form, find the integrating factor, and solve.

Exam Tip: Standard linear differential equations are very common in board exams; practice identifying \(P\) and \(Q\) accurately.

Question 8. Solve \(x \frac{dy}{dx} + y + (x + 2)x^2 = 0\)
Answer: Rewriting as:
\(x \frac{dy}{dx} + y = -x^2(x + 2)\)
\(\frac{dy}{dx} + \frac{1}{x} y = -x(x + 2) = -x^2 - 2x\)
Here, \(P = \frac{1}{x}\) and \(Q = -x^2 - 2x\).
\(\text{I.F.} = e^{\int \frac{1}{x} dx} = x\).
Solution:
\(y \cdot x = \int (-x^2 - 2x)x \, dx + C = \int (-x^3 - 2x^2) dx + C\)
\(xy = -\frac{x^4}{4} - \frac{2x^3}{3} + C\)
\(y = -\frac{x^3}{4} - \frac{2x^2}{3} + \frac{C}{x}\)
In simple words: Divide by \(x\) to form a linear differential equation, compute the integrating factor \(x\), and integrate.

Exam Tip: Always distribute \(x\) into the polynomial \(Q\) before integrating to make term-by-term integration easier.

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 standard form \(\frac{dy}{dx} + Py = Q\), we get \(P = -\frac{1}{x}\).
\(\text{I.F.} = e^{\int P dx} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = e^{\log(x^{-1})} = \frac{1}{x}\).
In simple words: Identify \(P\) as the coefficient of \(y\) and compute \(e^{\int P dx}\).

Exam Tip: Include the negative sign when identifying \(P\); missing it is a common error.

LEVEL II

Question 1. Solve \(\frac{dy}{dx} + 2y \tan x = \sin x\)
Answer: Here, \(P = 2 \tan x\) and \(Q = \sin x\).
\(\text{I.F.} = e^{\int 2 \tan x dx} = e^{2 \log|\sec x|} = e^{\log(\sec^2 x)} = \sec^2 x\).
Solution:
\(y \cdot \sec^2 x = \int \sin x \cdot \sec^2 x dx + C = \int \tan x \sec x dx + C\)
\(y \sec^2 x = \sec x + C\)
\(y = \cos x + C \cos^2 x\)
In simple words: Find the integrating factor using logarithmic properties of trigonometric integrals, then integrate.

Exam Tip: \(\int \tan x \sec x dx = \sec x\) is a standard integration result.

Question 2. Solve \((1 + x^2)\frac{dy}{dx} + y = e^x(x + 1)^2\)
Answer: Dividing by \((1 + x^2)\):
\(\frac{dy}{dx} + \frac{1}{1 + x^2} y = \frac{e^x(x + 1)^2}{1 + x^2}\)
Here, \(P = \frac{1}{1 + x^2}\) and \(Q = \frac{e^x(x^2 + 2x + 1)}{1 + x^2}\).
\(\text{I.F.} = e^{\int \frac{1}{1 + x^2} dx} = e^{\tan^{-1} x}\).
Solution:
\(y \cdot e^{\tan^{-1} x} = \int \frac{e^x(x + 1)^2}{1 + x^2} e^{\tan^{-1} x} dx + C\)
In simple words: Put the equation in standard form, find the integrating factor \(e^{\tan^{-1} x}\), and evaluate the integral.

Question 3. Solve \(x \frac{dy}{dx} + y = x \log x\)
Answer: Dividing by \(x\):
\(\frac{dy}{dx} + \frac{1}{x} y = \log x\)
Here, \(P = \frac{1}{x}\) and \(Q = \log x\).
\(\text{I.F.} = e^{\int \frac{1}{x} dx} = x\).
Solution:
\(y \cdot x = \int (\log x)x \, dx + C\)
Using integration by parts on \(\int x \log x dx\):
\(\int x \log x dx = \log x \left(\frac{x^2}{2}\right) - \int \frac{1}{x} \left(\frac{x^2}{2}\right) dx = \frac{x^2}{2}\log x - \frac{x^2}{4}\)
Thus, \(xy = \frac{x^2}{2}\log x - \frac{x^2}{4} + C\)
In simple words: Convert to standard linear form, find integrating factor \(x\), and apply integration by parts.

Exam Tip: Use LIATE rule for integration by parts when integrating logarithmic functions.

LEVEL III

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

Exam Tip: \(\frac{1}{1 + \cos v} = \frac{1}{2}\sec^2(v/2)\) makes the integration step immediate.

Question 2. Solve \(ye^x dx = (y^3 + 2xe^x)dy\)
Answer: Rewriting as:
\(\frac{dx}{dy} = \frac{y^3 + 2xe^x}{ye^x}\) -- wait, let's rearrange properly:
\((y^3 + 2xe^x)dy - ye^x dx = 0 \implies ye^x \frac{dx}{dy} - 2xe^x = y^3\)
Divide by \(ye^x\):
\(\frac{dx}{dy} - \frac{2}{y} x = \frac{y^2}{e^x}\) -- wait, rewriting: \(ye^x dx = (y^3 + 2xe^x)dy \implies \frac{dx}{dy} - \frac{2}{y}x = y^2 e^{-x}\)? No, let's check: \(ye^x dx - 2xe^x dy = y^3 dy \implies \frac{ye^x dx - 2xe^x dy}{y^3} = dy\)...
Let's divide the original equation by \(y^2 e^x\):
\(\frac{y e^x dx - 2x e^x dy}{y^3} = y \, dy \implies d\left(\frac{x e^x}{y^2}\right) = y \, dy\)
Integrating both sides:
\(\frac{x e^x}{y^2} = \frac{y^2}{2} + C\)
\(2x e^x = y^4 + 2Cy^2\)
In simple words: Group terms to recognize an exact differential quotient rule for \(\frac{x e^x}{y^2}\).

Exam Tip: Recognizing exact differentials saves considerable time compared to standard integrating factor methods.

Question 3. Solve \(x^2 \frac{dy}{dx} = y(x + y)\)
Answer: Rewriting 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 \(y = vx \implies \frac{dy}{dx} = v + x \frac{dv}{dx}\).
\(v + x \frac{dv}{dx} = v + v^2\)
\(x \frac{dv}{dx} = v^2\)
Separating variables:
\(\frac{dv}{v^2} = \frac{dx}{x}\)
Integrating both sides:
\[ -\frac{1}{v}=\log|x|+C \]
Substitute \(v = \frac{y}{x}\):
\(-\frac{x}{y} = \log|x| + C \implies y(\log|x| + C) + x = 0\)
In simple words: Substitute \(y = vx\), simplify to get a separable equation in \(v\) and \(x\), integrate, and substitute back.

Exam Tip: Homogeneous equations always reduce to variable separable form after the substitution \(y = vx\).

Question 4. Solve \(\frac{dy}{dx} + \frac{4x}{x^2 + 1}y = \frac{1}{(x^2 + 1)^3}\)
Answer: This is a linear differential equation where \(P = \frac{4x}{x^2 + 1}\) and \(Q = \frac{1}{(x^2 + 1)^3}\).
\(\text{I.F.} = e^{\int \frac{4x}{x^2 + 1} dx} = e^{2\log(x^2 + 1)} = e^{\log((x^2 + 1)^2)} = (x^2 + 1)^2\).
Solution:
\(y \cdot (x^2 + 1)^2 = \int \frac{1}{(x^2 + 1)^3} \cdot (x^2 + 1)^2 dx + C\)
\(y(x^2 + 1)^2 = \int \frac{1}{x^2 + 1} dx + C\)
\(y(x^2 + 1)^2 = \tan^{-1} x + C\)
In simple words: Find the integrating factor using power rules of logarithms, then integrate directly.

Question 5. Solve the differential equation \((x + 2y^2)\frac{dy}{dx} = y\); given that when \(x = 2, y = 1\)
Answer: Rewriting as:
\(\frac{dx}{dy} = \frac{x + 2y^2}{y} = \frac{x}{y} + 2y\)
\(\frac{dx}{dy} - \frac{1}{y} x = 2y\)
This is a linear differential equation in \(x\): \(\frac{dx}{dy} + Px = Q$, where \(P = -\frac{1}{y}\) and \(Q = 2y\).
\(\text{I.F.} = e^{\int -\frac{1}{y} dy} = e^{-\log y} = \frac{1}{y}\).
Solution:
\(x \cdot \left(\frac{1}{y}\right) = \int 2y \cdot \left(\frac{1}{y}\right) dy + C\)
\(\frac{x}{y} = \int 2 dy + C = 2y + C\)
\(x = 2y^2 + Cy\)
Using the given condition \(x = 2\) when \(y = 1\):
\(2 = 2(1)^2 + C(1) \implies 2 = 2 + C \implies C = 0\)
Therefore, the particular solution is \(x = 2y^2\).
In simple words: Convert to a linear equation in \(x\), find the integrating factor, integrate, and use the given point to find \(C\).

Exam Tip: Always check whether to solve for \(\frac{dy}{dx}\) or \(\frac{dx}{dy}\) based on which variable's coefficient is simpler.

Questions for self evaluation

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

Exam Tip: Trigonometric, exponential, or logarithmic functions of derivatives make the degree not defined.

Question 2. Form the differential equation representing the family of ellipses having foci on \(x\)-axis and centre at origin.
Answer: The equation of the family of ellipses is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) and \(b\) are two arbitrary constants.
Differentiating with respect to \(x\):
\(\frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0 \implies \frac{x}{a^2} + \frac{y}{b^2}y' = 0\) -- (1)
Differentiating again with respect to \(x\):
\(\frac{1}{a^2} + \frac{1}{b^2}\left[y y'' + (y')^2\right] = 0\) -- (2)
Solving equations (1) and (2) for \(\frac{1}{a^2}\) and \(\frac{1}{b^2}\) and substituting back gives 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 twice to eliminate the two arbitrary constants \(a\) and \(b\).

Exam Tip: Two arbitrary constants require a second-order differential equation.

Question 3. Solve the differential equation \((\tan^{-1} y - x)dy = (1 + y^2)dx\), given that \(y = 0\) when \(x = 0\).
Answer: Rearranging:
\(\frac{dx}{dy} = \frac{\tan^{-1} y - x}{1 + y^2} = \frac{\tan^{-1} y}{1 + y^2} - \frac{x}{1 + y^2}\)
\(\frac{dx}{dy} + \frac{1}{1 + y^2} x = \frac{\tan^{-1} y}{1 + y^2}\)
This is a linear differential equation in \(x\). Here \(P = \frac{1}{1 + y^2}\) and \(Q = \frac{\tan^{-1} y}{1 + y^2}\).
\(\text{I.F.} = e^{\int \frac{1}{1 + y^2} dy} = e^{\tan^{-1} y}\).
Solution:
\(x \cdot e^{\tan^{-1} y} = \int \frac{\tan^{-1} y}{1 + y^2} e^{\tan^{-1} y} dy + C\)
Let \(\tan^{-1} y = t \implies \frac{1}{1 + y^2}dy = dt\).
\(x e^{\tan^{-1} y} = \int t e^t dt + C = (t - 1)e^t + C = (\tan^{-1} y - 1)e^{\tan^{-1} y} + C\)
\(x = \tan^{-1} y - 1 + C e^{-\tan^{-1} y}\)
Using the condition \(x = 0\) when \(y = 0\):
\(0 = \tan^{-1}(0) - 1 + C e^{-\tan^{-1}(0)} \implies 0 = 0 - 1 + C(1) \implies C = 1\)
Therefore, the particular solution is:
\(x = \tan^{-1} y - 1 + e^{-\tan^{-1} y}\)
In simple words: Convert to linear form in \(x\), use integrating factor, integrate by parts, and apply the initial condition.

Exam Tip: Substitution simplifies integrals involving inverse trigonometric functions significantly.

Question 4. Solve the differential equation \(x \, dy - y \, dx = \sqrt{x^2 + y^2} \, dx\)
Answer: This is identical to Question 5 of Level III (Homogeneous Equations).
Rearranging and substituting \(y = vx\) gives the solution:
\(y + \sqrt{x^2 + y^2} = C x^2\)
In simple words: Use homogeneous substitution \(y = vx\) to solve.

Exam Tip: Practice standard homogeneous forms thoroughly as they appear frequently.

Question 5. Solve the differential equation \(x \log x \frac{dy}{dx} + y = \frac{2}{x}\log x\)
Answer: Dividing by \(x \log x\):
\(\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x^2}\)
Here, \(P = \frac{1}{x \log x}\) and \(Q = \frac{2}{x^2}\).
\(\text{I.F.} = e^{\int \frac{1}{x \log x} dx} = e^{\log(\log x)} = \log x\).
Solution:
\(y \cdot \log x = \int \frac{2}{x^2} \log x \, dx + C\)
Using integration by parts on \(\int x^{-2} \log x dx\):
\(\int x^{-2} \log x dx = \log x \left(-\frac{1}{x}\right) - \int \frac{1}{x} \left(-\frac{1}{x}\right) dx = -\frac{\log x}{x} - \frac{1}{x}\)
Thus, \(y \log x = 2\left(-\frac{\log x + 1}{x}\right) + C\)
In simple words: Convert to standard linear form, find the integrating factor, and use integration by parts.

Exam Tip: Logarithmic integrating factors simplify nicely to algebraic functions of logarithms.

Question 6. Solve the differential equation \(x^2 dy + (y^2 + xy)dx = 0\), \(y(1) = 1\)
Answer: Rewriting as:
\(\frac{dy}{dx} = -\frac{y^2 + xy}{x^2}\)
This is a homogeneous differential equation. Put \(y = vx \implies \frac{dy}{dx} = v + x \frac{dv}{dx}\).
\(v + x \frac{dv}{dx} = -\frac{v^2 x^2 + vx^2}{x^2} = -(v^2 + v)\)
\(x \frac{dv}{dx} = -v^2 - 2v\)
Separating variables:
\(\frac{dv}{v(v + 2)} = -\frac{dx}{x}\)
Using partial fractions: \(\frac{1}{v(v+2)} = \frac{1}{2}\left(\frac{1}{v} - \frac{1}{v+2}\right)\).
\(\frac{1}{2}(\log|v| - \log|v + 2|) = -\log|x| + C\)
\(\log\left(\frac{v}{v + 2}\right) = -2\log|x| + C_1 = \log(x^{-2}) + C_1\)
\(\frac{v}{v + 2} = \frac{C_2}{x^2}\)
Substitute \(v = \frac{y}{x}\):
\(\frac{\frac{y}{x}}{\frac{y}{x} + 2} = \frac{C_2}{x^2} \implies \frac{y}{y + 2x} = \frac{C_2}{x^2}\)
Using \(y(1) = 1\) (i.e., \(x = 1, y = 1\)):
\(\frac{1}{1 + 2} = C_2 \implies C_2 = \frac{1}{3}\)
Therefore, the particular solution is:
\(\frac{y}{y + 2x} = \frac{1}{3x^2} \implies 3x^2 y = y + 2x\)
In simple words: Use homogeneous substitution, apply partial fractions for integration, and determine the constant using the initial condition.

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

Question 7. Show that the differential equation \(2y \cdot e^{x/y} dx + \left(y - 2x \cdot e^{x/y}\right)dy = 0\) is homogeneous and find its particular solution given that \(y(0) = 1\).
Answer: Rewriting as:
\(\frac{dx}{dy} = \frac{2x e^{x/y} - y}{2y e^{x/y}}\)
This is homogeneous in \(x\) and \(y\). Put \(x = vy \implies \frac{dx}{dy} = v + y \frac{dv}{dy}\).
\(v + y \frac{dv}{dy} = \frac{2vy e^v - y}{2y e^v} = v - \frac{1}{2e^v}\)
\(y \frac{dv}{dy} = -\frac{1}{2e^v}\)
Separating variables:
\(2e^v dv = -\frac{dy}{y}\)
Integrating both sides:
\(2e^v = -\log|y| + C\)
Substitute \(v = \frac{x}{y}\):
\(2e^{x/y} + \log|y| = C\)
Using \(y(0) = 1\) when \(x = 0\):
\(2e^0 + \log(1) = C \implies 2(1) + 0 = C \implies C = 2\)
Therefore, the particular solution is:
\(2e^{x/y} + \log|y| = 2\)
In simple words: Since terms feature \(\frac{x}{y}\), substitute \(x = vy\) and \(\frac{dx}{dy}\), then integrate.

Exam Tip: When terms contain \(\frac{x}{y}\) rather than \(\frac{y}{x}\), use the substitution \(x = vy\) and find \(\frac{dx}{dy}\).

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 where \(P = \cot x\) and \(Q = 2x + x^2 \cot x\).
\(\text{I.F.} = e^{\int \cot x dx} = e^{\log|\sin x|} = \sin x\).
Solution:
\(y \cdot \sin x = \int (2x + x^2 \cot x) \sin x \, dx + C\)
\(y \sin x = \int (2x \sin x + x^2 \cos x) dx + C\)
Notice that \(\frac{d}{dx}(x^2 \sin x) = 2x \sin x + x^2 \cos x\).
Thus, \(y \sin x = x^2 \sin x + C\)
Using the given condition \(y = 0\) when \(x = \frac{\pi}{2}\):
\(0 \cdot \sin\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) + C\)
\(0 = \frac{\pi^2}{4}(1) + C \implies C = -\frac{\pi^2}{4}\)
Therefore, the particular solution is:
\(y \sin x = x^2 \sin x - \frac{\pi^2}{4} \implies y = x^2 - \frac{\pi^2}{4}\csc x\)
In simple words: Find the integrating factor, recognize the product rule derivative on the right-hand side, integrate, and apply the boundary condition.

Exam Tip: Spotting product rule expansions inside integrals saves time compared to applying integration by parts repeatedly.