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Q15:
If $y = 500e^{7x} + 600e^{-7x}$, show that $\frac{d^2y}{dx^2} = 49y$
Solution :
Step 1: Establishing the Initial Function
The given function is:
$y = 500e^{7x} + 600e^{-7x}$ --- (Equation 1)
Step 2: First Order Differentiation
Differentiating Equation 1 with respect to $x$ using the Chain Rule $\frac{d}{dx}[e^{u(x)}] = e^{u(x)} \cdot \frac{du}{dx}$ [Per the Chain Rule of Differentiation]:
$\frac{dy}{dx} = 500 \cdot \frac{d}{dx}(e^{7x}) + 600 \cdot \frac{d}{dx}(e^{-7x})$
$\frac{dy}{dx} = 500(e^{7x} \cdot 7) + 600(e^{-7x} \cdot -7)$
$\frac{dy}{dx} = 3500e^{7x} - 4200e^{-7x}$ --- (Equation 2)
Step 3: Second Order Differentiation
Differentiating Equation 2 with respect to $x$ to obtain the second-order derivative:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(3500e^{7x} - 4200e^{-7x})$
$\frac{d^2y}{dx^2} = 3500 \cdot \frac{d}{dx}(e^{7x}) - 4200 \cdot \frac{d}{dx}(e^{-7x})$
$\frac{d^2y}{dx^2} = 3500(e^{7x} \cdot 7) - 4200(e^{-7x} \cdot -7)$
Step 4: Algebraic Simplification
Multiplying the terms:
$\frac{d^2y}{dx^2} = 24500e^{7x} + 29400e^{-7x}$
Step 5: Factorization and Verification
To demonstrate that $\frac{d^2y}{dx^2} = 49y$, we factor out $49$ from the expression obtained in Step 4:
$\frac{d^2y}{dx^2} = 49(500e^{7x} + 600e^{-7x})$
Comparing the expression within the parentheses to Equation 1, we observe:
$500e^{7x} + 600e^{-7x} = y$
Therefore:
$\frac{d^2y}{dx^2} = 49y$
Visual Representation of the Derivative Relationship
Final Solution: Since $\frac{d^2y}{dx^2} = 24500e^{7x} + 29400e^{-7x}$ and $y = 500e^{7x} + 600e^{-7x}$, it is proven that $\frac{d^2y}{dx^2} = 49y$.
More Questions from Class 12 Mathematics Continuity and Differentiability EXERCISE 5.7
- Q1: Find the second order derivatives of the functions $x^2 + 3x + 2$
- Q2: Find the second order derivatives of the functions $x^{20}$
- Q3: Find the second order derivatives of the functions $x \cdot \cos x$
- Q4: Find the second order derivatives of the functions $\log x$
- Q5: Find the second order derivatives of the functions. $x^3 \log x$
- Q6: Find the second order derivatives of the functions. $e^x \sin 5x$
- Q7: Find the second order derivatives of the functions. $e^{6x} \cos 3x$
- Q8: Find the second order derivatives of the functions. $\tan^{-1} x$
- Q9: Find the second order derivatives of the functions. $\log(\log x)$
- Q10: Find the second order derivatives of the functions. $\sin(\log x)$
- Q11: If $y = 5 \cos x - 3 \sin x$, prove that $\frac{d^2y}{dx^2} + y = 0$
- Q12: If $y = \cos^{-1} x$, Find $\frac{d^2y}{dx^2}$ in terms of $y$ alone.
- Q13: If $y = 3 \cos(\log x) + 4 \sin(\log x)$, show that $x^2 y_2 + xy_1 + y = 0$
- Q14: If $y = Ae^{mx} + Be^{nx}$, show that $\frac{d^2y}{dx^2} - (m + n)\frac{dy}{dx} + mny = 0$
- Q16: If $e^y (x + 1) = 1$, show that $\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2$
- Q17: If $y = (\tan^{-1} x)^2$, show that $(x^2 + 1)^2 y_2 + 2x(x^2 + 1)y_1 = 2$
CBSE Solutions for Class 12 Mathematics Continuity and Differentiability
- Continuity and Differentiability Class 12 EXERCISE 5.1
- Continuity and Differentiability Class 12 EXERCISE 5.2
- Continuity and Differentiability Class 12 EXERCISE 5.3
- Continuity and Differentiability Class 12 EXERCISE 5.4
- Continuity and Differentiability Class 12 EXERCISE 5.5
- Continuity and Differentiability Class 12 EXERCISE 5.6
Chapters in CBSE - Class 12 Mathematics
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