Maharashtra Board Class 12 Maths Part 1 Chapter 8 Differential Equation 8.1 Solutions

Question 1. Determine the order and degree of each of the following differential equations:
(i) \( \frac{d^2x}{dt^2} + \left(\frac{dx}{dt}\right)^2 + 8 = 0 \)
Answer:
Solution:
The given D.E. is \( \frac{d^2x}{dt^2} + \left(\frac{dx}{dt}\right)^2 + 8 = 0 \)
This D.E. has highest order derivative \( \frac{d^2x}{dt^2} \) with power 1.
\( \therefore \) the given D.E. is of order 2 and degree 1.
In simple words: The order of a differential equation is determined by the highest derivative present, which is the second derivative here (order 2). The degree is the power of that highest derivative term, which is 1 in this case.

🎯 Exam Tip: Always identify the highest order derivative first, then its power to correctly determine the order and degree. Ensure the equation is a polynomial in its derivatives before finding the degree.

(ii) \( \left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^2 = a^x \)
Answer:
Solution:
The given D.E. is \( \left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^2 = a^x \)
This D.E. has highest order derivative \( \frac{d^2y}{dx^2} \) with power 2.
\( \therefore \) the given D.E. is of order 2 and degree 2.
In simple words: For this equation, the highest derivative is the second derivative, making the order 2. The power of this highest derivative term is also 2, which gives its degree.

🎯 Exam Tip: Carefully check the powers of all derivative terms. The degree is the power of the *highest order* derivative, not necessarily the highest power overall.

(iii) \( \frac{d^4y}{dx^4} + \left[1 + \left(\frac{dy}{dx}\right)^2\right]^3 = 0 \)
Answer:
Solution:
The given D.E. is \( \frac{d^4y}{dx^4} + \left[1 + \left(\frac{dy}{dx}\right)^2\right]^3 = 0 \)
This D.E. has highest order derivative \( \frac{d^4y}{dx^4} \) with power 1.
\( \therefore \) the given D.E. is of order 4 and degree 1.
In simple words: Here, the fourth derivative is the highest, establishing the order as 4. Since this highest derivative term is raised to the power of 1, the degree is 1.

🎯 Exam Tip: Even if other derivative terms have higher powers, only the power of the *highest order* derivative determines the degree, assuming the equation is a polynomial in derivatives.

(iv) \( (y''')^2 + 2(y'')^2 + 6y' + 7y = 0 \)
Answer:
Solution:
The given D.E. is \( (y''')^2 + 2(y'')^2 + 6y' + 7y = 0 \)
This can be written as \( \left(\frac{d^3y}{dx^3}\right)^2 + 2\left(\frac{d^2y}{dx^2}\right)^2 + 6\frac{dy}{dx} + 7y = 0 \)
This D.E. has highest order derivative \( \frac{d^3y}{dx^3} \) with power 2.
\( \therefore \) the given D.E. is of order 3 and degree 2.
In simple words: The third derivative is the highest in this equation, setting the order to 3. Its power is 2, which determines the degree as 2.

🎯 Exam Tip: Remember that `y'`, `y''`, `y'''` notations correspond to first, second, and third derivatives respectively, and apply the rules consistently.

(v) \( \sqrt{1 + \frac{1}{\left(\frac{dy}{dx}\right)^2}} = \left(\frac{dy}{dx}\right)^{3/2} \)
Answer:
Solution:
The given D.E. is \( \sqrt{1 + \frac{1}{\left(\frac{dy}{dx}\right)^2}} = \left(\frac{dy}{dx}\right)^{3/2} \)
On squaring both sides, we get
\( 1 + \frac{1}{\left(\frac{dy}{dx}\right)^2} = \left(\frac{dy}{dx}\right)^3 \)
\( \implies \)
\( \left(\frac{dy}{dx}\right)^2 + 1 = \left(\frac{dy}{dx}\right)^5 \)
This D.E. has highest order derivative \( \frac{dy}{dx} \) with power 5.
\( \therefore \) the given D.E. is of order 1 and degree 5.
In simple words: To find the order and degree, first clear any fractional powers or radicals by algebraic manipulation. Once the equation is a polynomial in its derivatives, identify the highest derivative and its power.

🎯 Exam Tip: Always simplify the differential equation into a polynomial form with respect to its derivatives *before* determining the degree. Failing to do so can lead to incorrect degree identification.

(vi) \( \frac{dy}{dx} = 7 \frac{d^2y}{dx^2} \)
Answer:
Solution:
The given D.E. is \( \frac{dy}{dx} = 7 \frac{d^2y}{dx^2} \)
This D.E. has highest order derivative \( \frac{d^2y}{dx^2} \) with power 1.
\( \therefore \) the given D.E. is of order 2 and degree 1.
In simple words: In this equation, the second derivative is the highest, making the order 2. Since its power is 1, the degree is 1.

🎯 Exam Tip: Even if the highest order derivative appears on one side and a lower order on the other, the fundamental rules for order and degree remain the same.

(vii) \( \left(\frac{d^3y}{dx^3}\right)^{1/6} = 9 \)
Answer:
Solution:
The given D.E. is \( \left(\frac{d^3y}{dx^3}\right)^{1/6} = 9 \)
i.e., \( \frac{d^3y}{dx^3} = 9^6 \)
This D.E. has highest order derivative \( \frac{d^3y}{dx^3} \) with power 1.
\( \therefore \) the given D.E. is of order 3 and degree 1.
In simple words: For equations with fractional powers, raise both sides to an appropriate power to eliminate them and express the equation as a polynomial in derivatives before finding the order and degree.

🎯 Exam Tip: Fractional powers on derivative terms must be eliminated by raising the entire equation to an integer power before correctly identifying the degree.

Question 2. In each of the following examples, verify that the given function is a solution of the corresponding differential equation:

Answer:
Solution:
(i) \( xy = \log y + k \)
Differentiating w.r.t. x, we get
\( x\frac{dy}{dx} + y \times 1 = \frac{1}{y}\frac{dy}{dx} + 0 \)
\( \implies \)
\( x\frac{dy}{dx} + y = \frac{1}{y}\frac{dy}{dx} \)
\( \implies \)
\( x\frac{dy}{dx} - \frac{1}{y}\frac{dy}{dx} = -y \)
\( \implies \)
\( \left(x - \frac{1}{y}\right)\frac{dy}{dx} = -y \)
\( \implies \)
\( \left(\frac{xy - 1}{y}\right)\frac{dy}{dx} = -y \)
\( \implies \)
\( \frac{dy}{dx} = \frac{-y^2}{xy - 1} = \frac{y^2}{1 - xy} \), if \( xy \neq 1 \)
\( \implies \)
\( y' = \frac{y^2}{1 - xy} \), if \( xy \neq 1 \).
\( \implies \)
\( y'(1 - xy) = y^2 \)
Hence, \( xy = \log y + k \) is a solution of the D.E. \( y'(1 - xy) = y^2 \).
In simple words: To verify if a given function is a solution to a differential equation, differentiate the function implicitly (or explicitly) the required number of times and substitute the derivatives back into the differential equation to check if it holds true.

🎯 Exam Tip: Pay close attention to implicit differentiation when both x and y are involved. Algebraic manipulation is key to matching the derived equation with the given differential equation.

(ii) \( y = x^n \)
Differentiating twice w.r.t. x, we get
\( \frac{dy}{dx} = \frac{d}{dx}(x^n) = nx^{n-1} \)
and
\( \frac{d^2y}{dx^2} = \frac{d}{dx}(nx^{n-1}) = n(n-1)x^{n-2} \)
Now substitute these into the D.E.:
\( x^2\frac{d^2y}{dx^2} - nx\frac{dy}{dx} + ny \)
\( = x^2 \cdot n(n-1)x^{n-2} - nx \cdot nx^{n-1} + n \cdot x^n \)
\( = n(n-1)x^n - n^2x^n + nx^n \)
\( = (n^2 - n - n^2 + n)x^n = 0 \)
This shows that \( y = x^n \) is a solution of the D.E.
\( x^2\frac{d^2y}{dx^2} - nx\frac{dy}{dx} + ny = 0 \)
In simple words: For explicit solutions, calculate the first and second derivatives and directly substitute them, along with the original function, into the differential equation to confirm equality.

🎯 Exam Tip: Careful calculation of derivatives is essential. Double-check all algebraic steps when substituting and simplifying to avoid errors.

(iii) \( y = e^x \)
Differentiating w.r.t. x, we get
\( \frac{dy}{dx} = e^x = y \)
Hence, \( y = e^x \) is a solution of the D.E. \( \frac{dy}{dx} = y \).
In simple words: This involves a simple verification: find the first derivative of the given function and check if it equals the function itself, as stated in the differential equation.

🎯 Exam Tip: For straightforward equations, the verification process is direct. Ensure you understand basic derivative rules, especially for exponential functions.

(iv) \( y = 1 - \log x \)
Differentiating w.r.t. x, we get
\( \frac{dy}{dx} = \frac{d}{dx}(1) - \frac{d}{dx}(\log x) \)
\( = 0 - \frac{1}{x} = -\frac{1}{x} \)
Differentiating again w.r.t. x, we get
\( \frac{d^2y}{dx^2} = \frac{d}{dx}(-x^{-1}) = -(-1)x^{-2} = \frac{1}{x^2} \)
Now substitute into the D.E.:
\( x^2\frac{d^2y}{dx^2} = x^2 \cdot \frac{1}{x^2} = 1 \)
Hence, \( y = 1 - \log x \) is a solution of the D.E.
\( x^2\frac{d^2y}{dx^2} = 1 \)
In simple words: Calculate the first and second derivatives of the given logarithmic function and substitute them into the differential equation to see if the identity holds.

🎯 Exam Tip: Remember the derivative of \( \log x \) is \( \frac{1}{x} \). Be careful with signs and power rules when differentiating negative powers.

(v) \( y = ae^x + be^{-x} \)
Differentiating w.r.t. x, we get
\( \frac{dy}{dx} = a(e^x) + b(-e^{-x}) = ae^x - be^{-x} \)
Differentiating again w.r.t. x, we get
\( \frac{d^2y}{dx^2} = a(e^x) - b(-e^{-x}) \)
\( = ae^x + be^{-x} \)
\( = y \)
Hence, \( y = ae^x + be^{-x} \) is a solution of the D.E. \( \frac{d^2y}{dx^2} = y \).
In simple words: This verification involves finding the second derivative of the given linear combination of exponential functions and comparing it directly to the original function `y`.

🎯 Exam Tip: Differentiating exponential functions \( e^x \) and \( e^{-x} \) correctly is vital. Note that \( \frac{d}{dx}(e^{-x}) = -e^{-x} \) and \( \frac{d^2}{dx^2}(e^{-x}) = e^{-x} \).

(vi) \( ax^2 + by^2 = 5 \)
Differentiating w.r.t. x, we get
\( a(2x) + b\left(2y\frac{dy}{dx}\right) = 0 \)
\( \implies \)
\( ax + by\frac{dy}{dx} = 0 \)
\( \implies \)
\( ax = -by\frac{dy}{dx} \) ......(1)
Differentiating again w.r.t. x, we get
\( a \cdot 1 = -b\left[\left(y\frac{d}{dx}\left(\frac{dy}{dx}\right)\right) + \left(\frac{dy}{dx}\frac{d}{dx}(y)\right)\right] \)
\( \implies \)
\( a = -b\left[y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2\right] \) ......(2)
Dividing (1) by (2), we get
\( \frac{ax}{a} = \frac{-by\frac{dy}{dx}}{-b\left[y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2\right]} \)
\( \implies \)
\( x = \frac{y\frac{dy}{dx}}{y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2} \)
\( \implies \)
\( xy\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 = y\frac{dy}{dx} \)
Hence, \( ax^2 + by^2 = 5 \) is a solution of the D.E.
\( xy\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 = y\frac{dy}{dx} \)
In simple words: For implicitly defined functions, use implicit differentiation to find the first and second derivatives, then substitute these along with the original function into the differential equation to check for equality.

🎯 Exam Tip: Implicit differentiation requires careful application of the chain rule. It's often easier to isolate \( \frac{dy}{dx} \) from the first differentiation before calculating the second derivative to substitute it back. Keep track of all terms to avoid errors.