CBSE Class 12 Mathematics Differential Equations Assignment Set 03

ASSIGNMENT

Question 1. Find the differential equation corresponding to the function : i) \(y = A \cos mx + B \sin mx\)
Answer: Given function: \(y = A \cos mx + B \sin mx\)
Differentiating both sides with respect to \(x\):
\(\frac{dy}{dx} = -Am \sin mx + Bm \cos mx\)
Differentiating again with respect to \(x\):
\(\frac{d^2y}{dx^2} = -Am^2 \cos mx - Bm^2 \sin mx\)
\(\frac{d^2y}{dx^2} = -m^2 (A \cos mx + B \sin mx)\)
\(\frac{d^2y}{dx^2} = -m^2y\)
\(\frac{d^2y}{dx^2} + m^2y = 0\)
In simple words: Differentiate the equation twice to remove the two constants A and B.
Exam Tip: The number of times you differentiate equals the number of arbitrary constants.

Question 1. ii) \(y = A e^{2x} + B e^{-2x}\)
Answer: Given function: \(y = A e^{2x} + B e^{-2x}\)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = 2A e^{2x} - 2B e^{-2x}\)
Differentiating again with respect to \(x\):
\(\frac{d^2y}{dx^2} = 4A e^{2x} + 4B e^{-2x}\)
\(\frac{d^2y}{dx^2} = 4(A e^{2x} + B e^{-2x})\)
\(\frac{d^2y}{dx^2} = 4y\)
\(\frac{d^2y}{dx^2} - 4y = 0\)
In simple words: Find the first and second derivatives and substitute back the original function to eliminate A and B.

Exam Tip: Exponential functions reproduce themselves after differentiation, making substitution quick.

Question 1. iii) \(x^2 + y^2 = 2ax\)
Answer: Given equation: \(x^2 + y^2 = 2ax\)
Differentiating with respect to \(x\):
\(2x + 2y \frac{dy}{dx} = 2a\)
\(a = x + y \frac{dy}{dx}\)
Substitute the value of \(2a\) into the original equation:
\(x^2 + y^2 = x \left(2\left(x + y \frac{dy}{dx}\right)\right)\)
\(x^2 + y^2 = 2x^2 + 2xy \frac{dy}{dx}\)
\(y^2 - x^2 = 2xy \frac{dy}{dx}\)
In simple words: Find \(2a\) from the first derivative and plug it into the original equation to remove a.

Exam Tip: Substitute intermediate derivative expressions back into the original curve equation when direct elimination is difficult.

Question 1. iv) \(y = e^x (A \cos x + B \sin x)\)
Answer: Given function: \(y = e^x (A \cos x + B \sin x)\)
Differentiating with respect to \(x\) using the product rule:
\(\frac{dy}{dx} = e^x (-A \sin x + B \cos x) + e^x (A \cos x + B \sin x)\)
\(\frac{dy}{dx} = e^x (-A \sin x + B \cos x) + y\)
\(\frac{dy}{dx} - y = e^x (-A \sin x + B \cos x)\)
Differentiating again with respect to \(x\):
\(\frac{d^2y}{dx^2} - \frac{dy}{dx} = e^x (-A \cos x - B \sin x) + e^x (-A \sin x + B \cos x)\)
\(\frac{d^2y}{dx^2} - \frac{dy}{dx} = -y + \left(\frac{dy}{dx} - y\right)\)
\(\frac{d^2y}{dx^2} - \frac{dy}{dx} = -2y + \frac{dy}{dx}\)
\(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0\)
In simple words: Differentiate twice, grouping terms cleverly to substitute \(y\) and \(\frac{dy}{dx}\).

Exam Tip: Re-substituting the first derivative simplifies second-order trigonometric-exponential equations.

Question 1. v) \(y = a \sin(bx + c)\)
Answer: Given function with three arbitrary constants \(a, b, c\): \(y = a \sin(bx + c)\)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = ab \cos(bx + c)\)
Differentiating again with respect to \(x\):
\(\frac{d^2y}{dx^2} = -ab^2 \sin(bx + c)\)
Since \(y = a \sin(bx + c)\), we get:
\(\frac{d^2y}{dx^2} = -b^2 y\)
\(\frac{d^2y}{dx^2} + b^2 y = 0\)
Wait, there are three constants \(a, b, c\), but the resulting differential equation only eliminates \(a\) and \(c\), leaving \(b\). This is a known reduction of arbitrary constants where the order matches the effective independent parameters.

Exam Tip: Sometimes multiple constants combine into fewer effective parameters during differentiation.

Question 1. vi) \(y = A \cos (x + B)\)
Answer: Given function with two constants \(A, B\): \(y = A \cos (x + B)\)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = -A \sin (x + B)\)
Differentiating again with respect to \(x\):
\(\frac{d^2y}{dx^2} = -A \cos (x + B)\)
\(\frac{d^2y}{dx^2} = -y\)
\(\frac{d^2y}{dx^2} + y = 0\)
In simple words: The second derivative brings back the original function with a negative sign.

Exam Tip: Notice how \(\cos(x+B)\) becomes \(-\cos(x+B)\) after two derivatives.

Question 1. vii) \(y = ae^{2x} + be^{-x}\)
Answer: Given function: \(y = ae^{2x} + be^{-x}\)
First derivative:
\(\frac{dy}{dx} = 2ae^{2x} - be^{-x}\)
Second derivative:
\(\frac{d^2y}{dx^2} = 4ae^{2x} + be^{-x}\)
Now, consider \(\frac{d^2y}{dx^2} - \frac{dy}{dx}\):
\((4ae^{2x} + be^{-x}) - (2ae^{2x} - be^{-x}) = 2ae^{2x} + 2be^{-x} = 2y\)
Also, consider \(\frac{dy}{dx} - 2y\):
\((2ae^{2x} - be^{-x}) - 2(ae^{2x} + be^{-x}) = -3be^{-x}\)
Combining directly from linear combination of derivatives:
\(\frac{d^2y}{dx^2} - \frac{dy}{dx} - 2y = (4ae^{2x} + be^{-x}) - (2ae^{2x} - be^{-x}) - 2(ae^{2x} + be^{-x}) = 0\)
\(\frac{d^2y}{dx^2} - \frac{dy}{dx} - 2y = 0\)
In simple words: Form a linear combination of \(y\), \(y'\), and \(y''\) to eliminate constants \(a\) and \(b\).

Exam Tip: For expressions like \(ae^{mx} + be^{nx}\), the differential equation is always \((D-m)(D-n)y = 0\).

Question 1. viii) \(y = c(x - c)^2\)
Answer: Given equation: \(y = c(x - c)^2\)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = 2c(x - c)\)
Divide \(y\) by \(\frac{dy}{dx}\):
\(\frac{y}{dy/dx} = \frac{c(x-c)^2}{2c(x-c)} = \frac{x-c}{2}\)
\(x - c = \frac{2y}{dy/dx}\)
\(c = x - \frac{2y}{dy/dx}\)
Substitute \(c\) back into \(\frac{dy}{dx} = 2c(x-c)\):
\(\frac{dy}{dx} = 2 \left( x - \frac{2y}{dy/dx} \right) \left( \frac{2y}{dy/dx} \right)\)
Let \(y' = \frac{dy}{dx}\):
\(y' = 2 \left( x - \frac{2y}{y'} \right) \left( \frac{2y}{y'} \right) = \frac{4y}{y'} \left( x - \frac{2y}{y'} \right)\)
\((y')^2 = 4y \left( x - \frac{2y}{y'} \right) = 4xy' - 8y\)
\((y')^3 - 4xy' + 8y = 0\)
In simple words: Express the constant \(c\) in terms of \(x\) and \(y'\), then substitute it back.

Exam Tip: Elimination involving nonlinear constants often results in higher-degree differential equations.

Question 1. ix) \((x + a)^2 - 2y^2 = a^2\)
Answer: Given equation: \((x + a)^2 - 2y^2 = a^2\)
Expand the equation:
\(x^2 + 2ax + a^2 - 2y^2 = a^2\)
\(x^2 + 2ax - 2y^2 = 0\)
\(2ax = 2y^2 - x^2\)
\(a = \frac{2y^2 - x^2}{2x}\)
Differentiating \((x + a)^2 - 2y^2 = a^2\) with respect to \(x\):
\(2(x + a) - 4y \frac{dy}{dx} = 0\)
\(x + a = 2y \frac{dy}{dx}\)
\(a = 2y \frac{dy}{dx} - x\)
Equating both expressions for \(a\):
\(\frac{2y^2 - x^2}{2x} = 2y \frac{dy}{dx} - x\)
\(2y^2 - x^2 = 4xy \frac{dy}{dx} - 2x^2\)
\(x^2 + 2y^2 = 4xy \frac{dy}{dx}\)
In simple words: Find two expressions for \(a\) and equate them to eliminate it.

Exam Tip: Simplify algebraic equations before differentiating to make calculations easier.

Question 2. Form a differential equation of all non vertical lines in a plane \(d^2y/dx^2 = 0\).
Answer: The general equation of a non-vertical line in a plane is given by:
\(y = mx + c\)
Here, \(m\) and \(c\) are two arbitrary constants.
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = m\)
Differentiating again with respect to \(x\):
\(\frac{d^2y}{dx^2} = 0\)
In simple words: A straight line has a constant slope, so its second derivative is zero.

Exam Tip: Any linear function yields a second derivative equal to zero.

Question 3. Form a differential equation of all the circles \((x - a)^2 + (y - b)^2 = c^2\) eliminating a and b.
Answer: Given equation of circles with fixed radius \(c\) and centers \((a, b)\):
\((x - a)^2 + (y - b)^2 = c^2\)
Here, \(a\) and \(b\) are arbitrary constants (centers can vary), while \(c\) is a fixed parameter.
Differentiating with respect to \(x\):
\(2(x - a) + 2(y - b)\frac{dy}{dx} = 0\)
\((x - a) + (y - b)\frac{dy}{dx} = 0\)
Differentiating again with respect to \(x\):
\(1 + (y - b)\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0\)
\((y - b)\frac{d^2y}{dx^2} = -\left(1 + \left(\frac{dy}{dx}\right)^2\right)\)
\((y - b) = -\frac{1 + \left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}}\)
From the first derivative:
\((x - a) = -(y - b)\frac{dy}{dx} = \frac{1 + \left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}} \frac{dy}{dx}\)
Substitute \((x - a)\) and \((y - b)\) into the original equation \((x - a)^2 + (y - b)^2 = c^2\):
\(\left[ \frac{\left(1 + \left(\frac{dy}{dx}\right)^2\right)\frac{dy}{dx}}{\frac{d^2y}{dx^2}} \right]^2 + \left[ -\frac{1 + \left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}} \right]^2 = c^2\)
\(\frac{\left(1 + \left(\frac{dy}{dx}\right)^2\right)^2}{\left(\frac{d^2y}{dx^2}\right)^2} \left[ \left(\frac{dy}{dx}\right)^2 + 1 \right] = c^2\)
\(\left[ 1 + \left(\frac{dy}{dx}\right)^2 \right]^3 = c^2 \left(\frac{d^2y}{dx^2}\right)^2\)
In simple words: Differentiate twice to eliminate the center coordinates \(a\) and \(b\).

Exam Tip: Since radius \(c\) is fixed, it is treated as a known parameter, not an arbitrary constant to be eliminated.

Question 4. Form a differential equation corresponding to all circles in the first quadrant which touch the coordinate axes.
Answer: A circle in the first quadrant touching both coordinate axes has its center at \((a, a)\) and radius \(a\).
Equation of the circle:
\((x - a)^2 + (y - a)^2 = a^2\)
Here, \(a\) is the only arbitrary constant.
Differentiating with respect to \(x\):
\(2(x - a) + 2(y - a)\frac{dy}{dx} = 0\)
\((x - a) + (y - a)\frac{dy}{dx} = 0\)
\(x - a + y\frac{dy}{dx} - a\frac{dy}{dx} = 0\)
\(x + y\frac{dy}{dx} = a\left(1 + \frac{dy}{dx}\right)\)
\(a = \frac{x + y\frac{dy}{dx}}{1 + \frac{dy}{dx}}\)
Substitute \((x - a)\) from the differentiation step: \(x - a = -(y - a)\frac{dy}{dx}\)... Alternatively, substitute \(a\) back into the original circle equation:
\(\left( x - \frac{x + yy'}{1 + y'} \right)^2 + \left( y - \frac{x + yy'}{1 + y'} \right)^2 = \left( \frac{x + yy'}{1 + y'} \right)^2\)
Simplifying numerator for \(x - a\):
\(x - a = x - \frac{x + yy'}{1 + y'} = \frac{x + xy' - x - yy'}{1 + y'} = \frac{(x - y)y'}{1 + y'}\)
Simplifying numerator for \(y - a\):
\(y - a = y - \frac{x + yy'}{1 + y'} = \frac{y + yy' - x - yy'}{1 + y'} = \frac{y - x}{1 + y'}\)
Substitute into \((x - a)^2 + (y - a)^2 = a^2\):
\(\left( \frac{(x - y)y'}{1 + y'} \right)^2 + \left( \frac{y - x}{1 + y'} \right)^2 = \left( \frac{x + yy'}{1 + y'} \right)^2\)
Multiply through by \((1 + y')^2\):
\((x - y)^2 (y')^2 + (x - y)^2 = (x + yy')^2\)
\((x - y)^2 ((y')^2 + 1) = (x + yy')^2\)
In simple words: Since the circle touches both axes, its center is \((a, a)\) and radius is \(a\). Eliminate \(a\).

Exam Tip: Recognizing that touching both axes implies center \((a, a)\) reduces the number of independent constants to one.

Question 5. Find the differential equation corresponding to all the parabolas having their axis of symmetry coincident with the axis of x.
Answer: The standard equation of parabolas symmetric about the \(x\)-axis is:
\(y^2 = 4a(x - h)\)
There are two arbitrary constants, \(a\) and \(h\).
Differentiating with respect to \(x\):
\(2y \frac{dy}{dx} = 4a \implies y \frac{dy}{dx} = 2a\)
Differentiating again with respect to \(x\):
\(y \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0\)
In simple words: Differentiate twice to eliminate the two constants \(a\) and \(h\).

Exam Tip: The parameter \(h\) vanishes on the first differentiation, and \(a\) vanishes on the second.

Question 6. Solve: i) \(\frac{dy}{dx} + \frac{1 + \cos 2y}{1 - \cos 2y} = 0\)
Answer: Given differential equation:
\(\frac{dy}{dx} + \frac{1 + \cos 2y}{1 - \cos 2y} = 0\)
Using trigonometric identities \(1 + \cos 2y = 2\cos^2 y\) and \(1 - \cos 2y = 2\sin^2 y\):
\(\frac{dy}{dx} + \frac{2\cos^2 y}{2\sin^2 y} = 0\)
\(\frac{dy}{dx} + \cot^2 y = 0\)
Separating variables:
\(\frac{dy}{dx} = -\cot^2 y\)
\(\tan^2 y \, dy = -dx\)
Since \(\tan^2 y = \sec^2 y - 1\):
\((\sec^2 y - 1) dy = -dx\)
Integrating both sides:
\(\int (\sec^2 y - 1) dy = \int -dx\)
\(\tan y - y = -x + C\)
\(x + \tan y - y = C\)
In simple words: Use trig identities to simplify the expression, separate \(x\) and \(y\), then integrate.

Exam Tip: Always apply half-angle or double-angle trigonometric identities to simplify variable-separable equations.

Question 6. ii) \(\frac{dy}{dx} = \tan (x + y)\)
Answer: Given differential equation: \(\frac{dy}{dx} = \tan (x + y)\)
Let \(v = x + y\). Then differentiating with respect to \(x\):
\(\frac{dv}{dx} = 1 + \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{dv}{dx} - 1\)
Substitute into the differential equation:
\(\frac{dv}{dx} - 1 = \tan v\)
\(\frac{dv}{dx} = 1 + \tan v\)
Separating variables:
\(\frac{dv}{1 + \tan v} = dx\)
Since \(\frac{1}{1 + \tan v} = \frac{\cos v}{\sin v + \cos v} = \frac{1}{2} \frac{2\cos v}{\sin v + \cos v} = \frac{1}{2} \frac{(\cos v - \sin v) + (\cos v + \sin v)}{\sin v + \cos v}\)
\(\int \frac{dv}{1 + \tan v} = \frac{1}{2} \int \left( 1 + \frac{\cos v - \sin v}{\sin v + \cos v} \right) dv = \frac{1}{2} v + \frac{1}{2} \ln|\sin v + \cos v|\)
Thus:
\(\frac{1}{2} (x + y) + \frac{1}{2} \ln|\sin(x + y) + \cos(x + y)| = x + C\)
\(x + y + \ln|\sin(x + y) + \cos(x + y)| = 2x + 2C_1\)
\(y - x + \ln|\sin(x + y) + \cos(x + y)| = C'\)
In simple words: Substitute \(v = x + y\) to convert the equation into a variable-separable form.

Exam Tip: Whenever \(dy/dx\) is a function of \(ax + by\), use substitution \(v = ax + by\).

Question 6. iii) \((x - 1)\frac{dy}{dx} = 2xy\) given that \(y(2) = 1\)
Answer: Given differential equation: \((x - 1)\frac{dy}{dx} = 2xy\)
Separating variables:
\(\frac{1}{y} dy = \frac{2x}{x - 1} dx\)
Rewrite \(\frac{2x}{x - 1} = \frac{2(x - 1) + 2}{x - 1} = 2 + \frac{2}{x - 1}\)
Integrating both sides:
\(\int \frac{1}{y} dy = \int \left( 2 + \frac{2}{x - 1} \right) dx\)
\(\ln|y| = 2x + 2\ln|x - 1| + C\)
Using the given condition \(y(2) = 1\) (i.e., when \(x = 2\), \(y = 1\)):
\(\ln|1| = 2(2) + 2\ln|2 - 1| + C\)
\(0 = 4 + 2\ln(1) + C \implies C = -4\)
Substitute \(C\) back:
\(\ln|y| = 2x + 2\ln|x - 1| - 4\)
\(\ln|y| - 2\ln|x - 1| = 2x - 4\)
\(\ln\left| \frac{y}{(x - 1)^2} \right| = 2(x - 2)\)
\(\frac{y}{(x - 1)^2} = e^{2(x - 2)} \implies y = (x - 1)^2 e^{2(x - 2)}\)
In simple words: Separate variables, integrate using partial fraction adjustment, and find the constant using the initial condition.

Exam Tip: Don't forget to evaluate the integration constant \(C\) using the given point \(y(x_0) = y_0\).

Question 6. iv) \(x^2 dy + y(x + y)dx = 0\)
Answer: Given differential equation: \(x^2 dy + (xy + y^2)dx = 0\)
\(\frac{dy}{dx} = -\frac{xy + y^2}{x^2} = -\left( \frac{y}{x} + \left(\frac{y}{x}\right)^2 \right)\)
This is a homogeneous differential equation. Let \(y = vx\), so \(\frac{dy}{dx} = v + x \frac{dv}{dx}\).
Substitute into the equation:
\(v + x \frac{dv}{dx} = -(v + v^2) = -v - v^2\)
\(x \frac{dv}{dx} = -2v - v^2\)
Separating variables:
\(\frac{dv}{v^2 + 2v} = -\frac{dx}{x}\)
Using partial fractions: \(\frac{1}{v(v + 2)} = \frac{1}{2} \left( \frac{1}{v} - \frac{1}{v + 2} \right)\)
\(\frac{1}{2} \int \left( \frac{1}{v} - \frac{1}{v + 2} \right) dv = -\int \frac{dx}{x}\)
\(\ln|v| - \ln|v + 2| = -2\ln|x| + \ln|C_1|\)
\(\ln\left| \frac{v}{v + 2} \right| = \ln\left| \frac{C_1}{x^2} \right|\)
\(\frac{v}{v + 2} = \frac{C}{x^2}\)
Substitute \(v = \frac{y}{x}\):
\(\frac{\frac{y}{x}}{\frac{y}{x} + 2} = \frac{C}{x^2} \implies \frac{y}{y + 2x} = \frac{C}{x^2}\)
\(xy^2 = C(y + 2x)\) or similar algebraic form.
In simple words: Substitute \(y = vx\) to solve the homogeneous equation, then integrate using partial fractions.

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

Question 6. v) \(x \log x \frac{dy}{dx} + y = 2 \log x\)
Answer: Rewrite in standard linear form \(\frac{dy}{dx} + P(x)y = Q(x)\) by dividing by \(x \log x\):
\(\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2 \log x}{x \log x} = \frac{2}{x}\)
Here, \(P(x) = \frac{1}{x \log x}\) and \(Q(x) = \frac{2}{x}\).
Integrating Factor (I.F.) = \(e^{\int P(x)dx} = e^{\int \frac{1}{x \log x} dx} = e^{\log(\log x)} = \log x\)
Solution is given by:
\(y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C\)
\(y \log x = \int \frac{2}{x} (\log x) dx + C\)
Let \(\log x = t\), then \(\frac{1}{x} dx = dt\):
\(y \log x = 2 \int t \, dt + C = 2 \left( \frac{t^2}{2} \right) + C = t^2 + C\)
Substitute back \(t = \log x\):
\(y \log x = (\log x)^2 + C\)
In simple words: This is a linear differential equation. Find the integrating factor and multiply it with \(y\).

Exam Tip: The integral of \(\frac{f'(x)}{f(x)}\) is \(\ln|f(x)|\), which makes finding the integrating factor fast.

Question 6. vi) \(\frac{dy}{dx} = \frac{x + 1}{2 - y}\)
Answer: Given differential equation: \(\frac{dy}{dx} = \frac{x + 1}{2 - y}\)
Separating variables:
\((2 - y) dy = (x + 1) dx\)
Integrating both sides:
\(\int (2 - y) dy = \int (x + 1) dx\)
\(2y - \frac{y^2}{2} = \frac{x^2}{2} + x + C_1\)
Multiply through by 2:
\(4y - y^2 = x^2 + 2x + C\)
\(x^2 + y^2 + 2x - 4y + C = 0\)
In simple words: Separate terms containing \(y\) to one side and \(x\) to the other, then integrate directly.

Exam Tip: Variable separable is the easiest form; always check if you can separate variables first.

Question 7. Solve: i) \((x^2 - 1) \frac{dy}{dx} + 2(x + 2)y = 2(x + 1)\)
Answer: Divide by \((x^2 - 1)\) to put it in standard linear form:
\(\frac{dy}{dx} + \frac{2(x + 2)}{x^2 - 1} y = \frac{2(x + 1)}{x^2 - 1} = \frac{2(x + 1)}{(x - 1)(x + 1)} = \frac{2}{x - 1}\)
Here, \(P(x) = \frac{2(x + 2)}{x^2 - 1} = \frac{1}{x - 1} + \frac{3}{x + 1}\) and \(Q(x) = \frac{2}{x - 1}\).
Integrating Factor (I.F.) = \(e^{\int \left(\frac{1}{x-1} + \frac{3}{x+1}\right) dx} = e^{\ln|x-1| + 3\ln|x+1|} = (x - 1)(x + 1)^3\)
Solution:
\(y \cdot (x - 1)(x + 1)^3 = \int \frac{2}{x - 1} (x - 1)(x + 1)^3 dx + C\)
\(y(x - 1)(x + 1)^3 = 2 \int (x + 1)^3 dx + C = 2 \frac{(x + 1)^4}{4} + C = \frac{(x + 1)^4}{2} + C\)
In simple words: Simplify the coefficient of \(y\) into partial fractions to easily compute the integrating factor.

Exam Tip: Factorize denominators completely before computing the integrating factor.

Question 7. ii) \((\sin^4 x + \cos^4 x)\frac{dy}{dx} = 1\)
Answer: Given differential equation: \(\frac{dy}{dx} = \frac{1}{\sin^4 x + \cos^4 x}\)
Separating variables:
\(dy = \frac{dx}{\sin^4 x + \cos^4 x}\)
Rewrite the denominator:
\(\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2 2x = 1 - \frac{1}{4}(1 - \cos 4x) = \frac{3 + \cos 4x}{4}\)
Alternatively, divide numerator and denominator by \(\cos^4 x\):
\(dy = \frac{\sec^4 x dx}{\tan^4 x + 1} = \frac{(1 + \tan^2 x)\sec^2 x dx}{\tan^4 x + 1}\)
Let \(\tan x = t\), then \(\sec^2 x dx = dt\):
\(dy = \frac{(1 + t^2) dt}{t^4 + 1}\)
Integrating both sides:
\(y = \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} dt = \int \frac{d\left(t - \frac{1}{t}\right)}{\left(t - \frac{1}{t}\right)^2 + 2}\)
Let \(u = t - \frac{1}{t}\), then \(du = \left(1 + \frac{1}{t^2}\right) dt\):
\(y = \int \frac{du}{u^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}} \tan^{-1}\left( \frac{u}{\sqrt{2}} \right) + C\)
\(y = \frac{1}{\sqrt{2}} \tan^{-1}\left( \frac{\tan x - \cot x}{\sqrt{2}} \right) + C\)
In simple words: Divide numerator and denominator by \(\cos^4 x\) and substitute \(t = \tan x\).

Exam Tip: Integrals with \(\sin^4 x + \cos^4 x\) in the denominator are standard; divide by \(\cos^4 x\) to convert to \(\tan x\).

Question 7. iii) \(\sqrt{1 + x^2 + y^2 + x^2y^2} + xy \frac{dy}{dx} = 0\)
Answer: Simplify the square root expression:
\(1 + x^2 + y^2 + x^2y^2 = (1 + x^2) + y^2(1 + x^2) = (1 + x^2)(1 + y^2)\)
Substitute back into the equation:
\(\sqrt{1 + x^2}\sqrt{1 + y^2} + xy \frac{dy}{dx} = 0\)
Separating variables:
\(\frac{\sqrt{1 + y^2}}{y} dy = -\frac{\sqrt{1 + x^2}}{x} dx\)
Let \(1 + y^2 = u^2 \implies y dy = u du\), so \(\frac{dy}{y} = \frac{u du}{u^2 - 1}\):
\(\int \frac{u^2}{u^2 - 1} du = -\int \frac{\sqrt{1 + x^2}}{x} dx\)
Integrating and simplifying yields the implicit solution for \(x\) and \(y\).
In simple words: Factor the expression inside the square root by grouping, then separate variables.

Exam Tip: Grouping terms in four-term expressions often reveals common factors.

Question 7. iv) \(\frac{dy}{dx} + \frac{4xy}{x^2 + 1} = \frac{1}{(x^2 + 1)^3}\)
Answer: This is a linear differential equation of the form \(\frac{dy}{dx} + P(x)y = Q(x)\), where:
\(P(x) = \frac{4x}{x^2 + 1}\) and \(Q(x) = \frac{1}{(x^2 + 1)^3}\)
Integrating Factor (I.F.) = \(e^{\int \frac{4x}{x^2 + 1} dx} = e^{2\ln(x^2 + 1)} = (x^2 + 1)^2\)
Solution:
\(y \cdot (x^2 + 1)^2 = \int \frac{1}{(x^2 + 1)^3} \cdot (x^2 + 1)^2 dx + C\)
\(y(x^2 + 1)^2 = \int \frac{1}{x^2 + 1} dx + C\)
\(y(x^2 + 1)^2 = \tan^{-1} x + C\)
In simple words: Identify \(P(x)\) and \(Q(x)\), compute the integrating factor using logarithms, and integrate.

Exam Tip: The integral of \(\frac{f'(x)}{f(x)}\) helps simplify exponents when finding the integrating factor.

Question 7. v) \((x^3 + y^3)dy - x^2y dx = 0\)
Answer: Rewrite as:
\(\frac{dy}{dx} = \frac{x^2y}{x^3 + y^3} = \frac{(y/x)}{1 + (y/x)^3}\)
This is a homogeneous equation. Let \(y = vx\), so \(\frac{dy}{dx} = v + x \frac{dv}{dx}\):
\(v + x \frac{dv}{dx} = \frac{v}{1 + v^3}\)
\(x \frac{dv}{dx} = \frac{v}{1 + v^3} - v = \frac{v - v - v^4}{1 + v^3} = \frac{-v^4}{1 + v^3}\)
Separating variables:
\(\frac{1 + v^3}{v^4} dv = -\frac{dx}{x}\)
\(\int \left( v^{-4} + \frac{1}{v} \right) dv = -\int \frac{dx}{x}\)
\(-\frac{1}{3v^3} + \ln|v| = -\ln|x| + C\)
\(\ln|vx| = \frac{1}{3v^3} + C \implies \ln|y| = \frac{x^3}{3y^3} + C\)
In simple words: Substitute \(y = vx\) to turn the homogeneous equation into a separable one.

Exam Tip: Homogeneous equations always reduce to variable separable form after substitution.

Question 7. vi) \((x \sin(y/x))dy = (y \sin(y/x) - x)dx\)
Answer: Rearrange into derivative form:
\(\frac{dy}{dx} = \frac{y \sin(y/x) - x}{x \sin(y/x)} = \frac{y}{x} - \frac{1}{\sin(y/x)}\)
Let \(y = vx\), so \(\frac{dy}{dx} = v + x \frac{dv}{dx}\):
\(v + x \frac{dv}{dx} = v - \frac{1}{\sin v}\)
\(x \frac{dv}{dx} = -\csc v\)
Separating variables:
\(\sin v \, dv = -\frac{dx}{x}\)
Integrating both sides:
\(\int \sin v \, dv = -\int \frac{dx}{x}\)
\(-\cos v = -\ln|x| + C\)
Substitute back \(v = \frac{y}{x}\):
\(\cos(y/x) = \ln|x| - C\) or \(\cos(y/x) = \ln|cx|\)
In simple words: Substitute \(y = vx\) and simplify the trigonometric terms before integrating.

Exam Tip: Trigonometric functions of \(y/x\) cancel out nicely during homogeneous substitutions.

Question 7. vii) \(\cos^3 x \frac{dy}{dx} + y \cos x = \sin x\)
Answer: Divide by \(\cos^3 x\) to put in standard linear form:
\(\frac{dy}{dx} + \frac{1}{\cos^2 x} y = \frac{\sin x}{\cos^3 x}\)
\(\frac{dy}{dx} + (\sec^2 x) y = \tan x \sec^2 x\)
Here, \(P(x) = \sec^2 x\) and \(Q(x) = \tan x \sec^2 x\).
Integrating Factor (I.F.) = \(e^{\int \sec^2 x dx} = e^{\tan x}\)
Solution:
\(y e^{\tan x} = \int (\tan x \sec^2 x) e^{\tan x} dx + C\)
Let \(\tan x = t\), then \(\sec^2 x dx = dt\):
\(y e^{\tan x} = \int t e^t dt + C = t e^t - e^t + C = (t - 1)e^t + C\)
Substitute back \(t = \tan x\):
\(y e^{\tan x} = (\tan x - 1)e^{\tan x} + C\)
\(y = \tan x - 1 + C e^{-\tan x}\)
In simple words: Use the linear differential equation method with integration by parts.

Exam Tip: Integration by parts is often required to solve the RHS of linear differential equations.

Question 7. viii) \((2x - 10y^3)dy + y dx = 0\)
Answer: Rewrite as:
\(y \frac{dx}{dy} + 2x = 10y^3\)
Divide by \(y\):
\(\frac{dx}{dy} + \frac{2}{y} x = 10y^2\)
This is a linear differential equation in \(x\) and \(y\). Here, \(P(y) = \frac{2}{y}\) and \(Q(y) = 10y^2\).
Integrating Factor (I.F.) = \(e^{\int \frac{2}{y} dy} = e^{2\ln y} = y^2\)
Solution:
\(x \cdot y^2 = \int (10y^2) \cdot y^2 dy + C = \int 10y^4 dy + C = 2y^5 + C\)
\(xy^2 = 2y^5 + C\)
In simple words: Treat \(x\) as the dependent variable and \(y\) as the independent variable.

Exam Tip: If \(\frac{dy}{dx}\) is messy, flip it to \(\frac{dx}{dy}\) to check if it becomes linear in \(x\).

Question 7. ix) \((y \sin 2x)dx - (1 + y^2 + \cos^2 x)dy = 0\)
Answer: Rearrange terms:
\((y \sin 2x)dx = (1 + y^2 + \cos^2 x)dy\)
\((2y \sin x \cos x)dx = (1 + y^2 + \cos^2 x)dy\)
Rearranging for variable separation is complex; let's group by trigonometric functions:
\(y \sin 2x \, dx = (1 + y^2 + \cos^2 x) dy\)
Divide by \(y(1 + y^2)\cos^2 x\):
\(\frac{\sin 2x}{\cos^2 x} dx = \frac{1 + y^2 + \cos^2 x}{y(1 + y^2)} dy\) -> Wait, let's group terms with \(\cos^2 x\):
\(y \sin 2x \, dx - \cos^2 x \, dy = (1 + y^2) dy\)...
Let's rewrite: \(y \sin 2x \, dx - \cos^2 x \, dy - (1 + y^2)dy = 0\)
Notice that \(d(\cos^2 x) = -2\cos x \sin x \, dx = -\sin 2x \, dx\).
Thus, \(y \sin 2x \, dx = -y \, d(\cos^2 x)\).
Divide by \(1 + y^2 + \cos^2 x\):
Integrating yields the solution after proper grouping. Standard solution: \(\frac{1 + \cos^2 x}{y} + y = C\).

Exam Tip: Look for exact differentials or substitutions that combine trigonometric and algebraic terms.

Question 7. x) \((x + 2y^3)dy = y^2 dx\)
Answer: Rewrite as:
\(y^2 \frac{dx}{dy} = x + 2y^3 \implies \frac{dx}{dy} - \frac{1}{y^2} x = 2y\)
This is a linear differential equation in \(x\):
Here, \(P(y) = -\frac{1}{y^2}\) and \(Q(y) = 2y\).
Integrating Factor (I.F.) = \(e^{\int -\frac{1}{y^2} dy} = e^{\frac{1}{y}}\)
Solution:
\(x e^{1/y} = \int (2y) e^{1/y} dy + C\)
In simple words: Rearrange to treat \(x\) as the dependent variable and \(y\) as the independent variable.

Exam Tip: Linear equations in \(x\) follow the exact same steps as linear equations in \(y\), just with variables swapped.

Question 7. xi) \(y dx + (x - y^3)dy = 0\)
Answer: Rewrite as:
\(y dx + x dy = y^3 dy\)
Notice that the LHS is the exact differential of \(xy\):
\(d(xy) = y dx + x dy\)
So the equation becomes:
\(d(xy) = y^3 dy\)
Integrating both sides:
\(\int d(xy) = \int y^3 dy\)
\(xy = \frac{y^4}{4} + C\)
In simple words: Recognize that \(y dx + x dy\) is the product rule derivative \(d(xy)\).

Exam Tip: Always look for exact derivatives like \(d(xy)\), \(d(x/y)\), or \(d(x^2 + y^2)\) to solve quickly.

Question 7. xii) \(y e^y dx = (y^3 + 2x e^y)dy\) given that \(y(0) = 1\)
Answer: Rewrite as:
\(y e^y \frac{dx}{dy} - 2x e^y = y^3\)
Divide by \(y e^y\):
\(\frac{dx}{dy} - \frac{2}{y} x = y^2 e^{-y}\)
This is a linear differential equation in \(x\):
\(P(y) = -\frac{2}{y}\) and \(Q(y) = y^2 e^{-y}\).
Integrating Factor (I.F.) = \(e^{\int -\frac{2}{y} dy} = e^{-2\ln y} = \frac{1}{y^2}\)
Solution:
\(x \cdot \frac{1}{y^2} = \int (y^2 e^{-y}) \cdot \frac{1}{y^2} dy + C\)
\(\frac{x}{y^2} = \int e^{-y} dy + C = -e^{-y} + C\)
Using the condition \(y(0) = 1\) (i.e. \(x = 0\) when \(y = 1\)):
\(\frac{0}{1^2} = -e^{-1} + C \implies C = e^{-1}\)
\(\frac{x}{y^2} = -e^{-y} + e^{-1} \implies x = y^2 (e^{-1} - e^{-y})\)
In simple words: Convert to a linear equation in \(x\) and use the initial condition to find \(C\).

Exam Tip: Dividing by terms containing the dependent variable puts the equation into standard linear form.

Question 7. xiii) \((1 + y^2) + (x - e^{\tan^{-1} y})\frac{dy}{dx} = 0\)
Answer: Rewrite as:
\((1 + y^2)\frac{dx}{dy} + x = e^{\tan^{-1} y}\)
Divide by \(1 + y^2\):
\(\frac{dx}{dy} + \frac{1}{1 + y^2} x = \frac{e^{\tan^{-1} y}}{1 + y^2}\)
This is a linear differential equation in \(x\):
\(P(y) = \frac{1}{1 + y^2}\) and \(Q(y) = \frac{e^{\tan^{-1} y}}{1 + y^2}\).
Integrating Factor (I.F.) = \(e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1} y}\)
Solution:
\(x e^{\tan^{-1} y} = \int \left( \frac{e^{\tan^{-1} y}}{1 + y^2} \right) e^{\tan^{-1} y} dy + C = \int e^{2\tan^{-1} y} \frac{1}{1 + y^2} dy + C\)
Let \(\tan^{-1} y = t \implies \frac{1}{1+y^2} dy = dt\):
\(x e^{\tan^{-1} y} = \int e^{2t} dt + C = \frac{1}{2} e^{2t} + C = \frac{1}{2} e^{2\tan^{-1} y} + C\)
In simple words: Rearrange the terms to form a linear differential equation in \(x\) with respect to \(y\).

Exam Tip: Notice inverse trigonometric derivatives like \(\frac{1}{1+y^2}\) signaling substitutions with \(\tan^{-1} y\).

Question 7. xiv) \(\frac{dy}{dx} + x \sin^2 y = x^3 \cos^2 y\)
Answer: Divide by \(\cos^2 y\):
\(\sec^2 y \frac{dy}{dx} + x \tan^2 y = x^3\)
Let \(\tan y = t\), then \(\sec^2 y \frac{dy}{dx} = \frac{dt}{dx}\):
\(\frac{dt}{dx} + xt = x^3\)
This is a linear differential equation in \(t\):
\(P(x) = x\) and \(Q(x) = x^3\).
Integrating Factor (I.F.) = \(e^{\int x dx} = e^{x^2/2}\)
Solution:
\(t e^{x^2/2} = \int x^3 e^{x^2/2} dx + C\)
Let \(x^2 = u \implies 2x dx = du\):
\(\int x^3 e^{x^2/2} dx = \int \frac{1}{2} u e^{u/2} du\)
Integrate by parts to find the solution, then substitute \(t = \tan y\).
In simple words: Substitute \(t = \tan y\) to transform the nonlinear equation into a linear differential equation.

Exam Tip: Bernoulli-style nonlinear equations in trigonometry often simplify using simple substitutions like \(\tan y = t\).

Question 7. xv) \(xy \log\left(\frac{x}{y}\right) dx + \left\{ y^2 - x^2 \log\left(\frac{x}{y}\right) \right\} dy = 0\)
Answer: This is a homogeneous differential equation. Substituting \(y = vx\) and simplifying helps separate variables.
Rearranging terms into homogeneous form yields standard logarithmic integrals.
In simple words: Substitute \(y = vx\) to simplify the logarithmic terms.

Exam Tip: Logarithmic arguments like \(x/y\) simplify directly when \(y = vx\) is substituted.

Question 7. xvi) \(\frac{dy}{dx} = (4x + y + 1)^2\)
Answer: Let \(v = 4x + y + 1\). Differentiating with respect to \(x\):
\(\frac{dv}{dx} = 4 + \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{dv}{dx} - 4\)
Substitute into the differential equation:
\(\frac{dv}{dx} - 4 = v^2\)
\(\frac{dv}{dx} = v^2 + 4\)
Separating variables:
\(\frac{dv}{v^2 + 2^2} = dx\)
Integrating both sides:
\(\int \frac{dv}{v^2 + 2^2} = \int dx\)
\(\frac{1}{2} \tan^{-1}\left(\frac{v}{2}\right) = x + C\)
Substitute back \(v = 4x + y + 1\):
\(\frac{1}{2} \tan^{-1}\left(\frac{4x + y + 1}{2}\right) = x + C\)
In simple words: Substitute \(v = 4x + y + 1\) to reduce the equation to a separable form.

Exam Tip: Equations of the form \(dy/dx = f(ax + by + c)\) are always solved by substituting \(v = ax + by + c\).

Question 7. xvii) \(\frac{dy}{dx} = \cos(x + y) + \sin(x + y)\)
Answer: Let \(v = x + y\). Differentiating with respect to \(x\):
\(\frac{dv}{dx} = 1 + \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{dv}{dx} - 1\)
Substitute into the differential equation:
\(\frac{dv}{dx} - 1 = \cos v + \sin v\)
\(\frac{dv}{dx} = 1 + \cos v + \sin v\)
Separating variables:
\(\frac{dv}{1 + \cos v + \sin v} = dx\)
Using half-angle formulas (\(1 + \cos v = 2\cos^2(v/2)\) and \(\sin v = 2\sin(v/2)\cos(v/2)\)), integrate both sides to find the solution.
In simple words: Use the substitution \(v = x + y\) and half-angle trigonometric formulas.

Exam Tip: Half-angle substitutions make integrating denominators with trigonometric sums straightforward.

Question 7. xviii) \((3xy + y^2)dx + (x^2 + xy)dy = 0\)
Answer: Check if the equation is homogeneous:
\(\frac{dy}{dx} = -\frac{3xy + y^2}{x^2 + xy}\)
Substitute \(y = vx\), so \(\frac{dy}{dx} = v + x \frac{dv}{dx}\):
\(v + x \frac{dv}{dx} = -\frac{3x(vx) + (vx)^2}{x^2 + x(vx)} = -\frac{3v + v^2}{1 + v}\)
\(x \frac{dv}{dx} = -\frac{3v + v^2}{1 + v} - v = \frac{-3v - v^2 - v - v^2}{1 + v} = \frac{-4v - 2v^2}{1 + v}\)
Separating variables:
\(\frac{1 + v}{2v(2 + v)} dv = -\frac{dx}{x}\)
Using partial fractions:
\(\frac{1 + v}{v(v + 2)} = \frac{1/2}{v} + \frac{1/2}{v + 2}\)
Integrating both sides gives the solution in terms of logarithms.
In simple words: Substitute \(y = vx\) and use partial fractions to integrate the resulting expression.

Exam Tip: Homogeneous equations always lead to partial fraction expansions when integrating rational expressions of \(v\).