Maharashtra Board Class 12 Maths Part 1 Chapter 8 Differential Equation 8.5 Solutions

Solve The Following Differential Equations.

Question 1. \( \frac{dy}{dx} + y = e^{-x} \)
Answer: Solution: \( \frac{dy}{dx} + y = e^{-x} \)........(1) This is the linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P=1 \) and \( Q=e^{-x} \)
Therefore, I.F. \( = e^{\int Pdx} = e^{\int 1 dx} = e^x \)
Therefore, the solution of (1) is given by \( y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dx + c \)
Therefore, \( y \cdot e^x = \int e^{-x} e^x dx + c \)
Therefore, \( e^x y = \int 1 dx + c \)
Therefore, \( e^x y = x + c \)
Therefore, \( y e^x = x + c \) This is the general solution.
In simple words: This problem involves solving a first-order linear differential equation by finding an integrating factor (I.F.) and then applying the general solution formula. The integrating factor helps simplify the equation into a form that can be easily integrated.

🎯 Exam Tip: Pay close attention to identifying P and Q correctly from the given differential equation, as errors here will propagate throughout the solution for the integrating factor and the final integral.

Question 2. \( \frac{dy}{dx} + y = 3 \)
Answer: Solution: \( \frac{dy}{dx} + y = 3 \) This is the linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = 1, Q = 3 \)
Therefore, I.F. \( = e^{\int Pdx} = e^{\int 1 dx} = e^x \)
Therefore, the solution of (1) is given by \( y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dx + c \)
Therefore, \( y e^x = \int 3 e^x dx + c = 3 e^x + c \)
Therefore, \( y e^x = 3 e^x + c \) This is the general solution.
In simple words: This problem is a straightforward application of solving a first-order linear differential equation. We calculate the integrating factor and then integrate to find the general solution.

🎯 Exam Tip: Remember to correctly identify the P and Q terms. A common mistake is to forget the constant of integration, 'c', which is crucial for a general solution.

Question 3. \( x \frac{dy}{dx} + 2y = x^2 \cdot \log x \).
Answer: Solution: \( x \frac{dy}{dx} + 2y = x^2 \cdot \log x \)
Therefore, \( \frac{dy}{dx} + \left( \frac{2}{x} \right) y = x \cdot \log x \)........(1) This is the linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{2}{x} \) and \( Q = x \cdot \log x \)
Therefore, I.F. \( = e^{\int Pdx} = e^{\int \frac{2}{x} dx} = e^{2 \int \frac{1}{x} dx} \) \( = e^{2 \log x} = e^{\log x^2} = x^2 \)
Therefore, the solution of (1) is given by \( y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dx + c \)
Therefore, \( y \cdot x^2 = \int (x \log x) x^2 dx + c \)
Therefore, \( x^2 y = \int x^3 \log x dx + c \) \( = (\log x) \int x^3 dx - \int \left[ \frac{d}{dx} (\log x) \int x^3 dx \right] dx + c \) \( = (\log x) \frac{x^4}{4} - \int \frac{1}{x} \cdot \frac{x^4}{4} dx + c \) \( = \frac{1}{4} x^4 \log x - \frac{1}{4} \int x^3 dx + c \)
Therefore, \( x^2 y = \frac{1}{4} x^4 \log x - \frac{x^4}{16} + c \)
Therefore, \( y \cdot x^2 = \frac{x^4 \log x}{4} - \frac{x^4}{16} + c \) This is the general solution.
In simple words: This problem involves a first-order linear differential equation that requires an initial step of division to bring it to the standard form. The integrating factor is calculated using the P term, and then integration by parts is used to solve the product of \( x^3 \) and \( \log x \).

🎯 Exam Tip: When the differential equation is not in the standard form \( \frac{dy}{dx} + Py = Q \), always divide by the coefficient of \( \frac{dy}{dx} \) first. Also, remember the integration by parts formula when dealing with products of functions.

Question 4. \( (x + y) \frac{dy}{dx} = 1 \)
Answer: Solution: \( (x + y) \frac{dy}{dx} = 1 \)
Therefore, \( \frac{dy}{dx} = x + y \)
Therefore, \( \frac{dy}{dx} - x = y \)
Therefore, \( \frac{dx}{dy} + (-1) x = y \)........(1) This is the linear differential equation of the form \( \frac{dx}{dy} + Px = Q \), where \( P = -1 \) and \( Q = y \)
Therefore, I.F. \( = e^{\int Pdy} = e^{\int -1 dy} = e^{-y} \)
Therefore, the solution of (1) is given by \( x \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dy + c \)
Therefore, \( x e^{-y} = \int y e^{-y} dy + c \)
Therefore, \( x e^{-y} = y \int e^{-y} dy - \int \left[ \frac{d}{dy} (y) \int e^{-y} dy \right] dy + c \) \( = y \left( \frac{e^{-y}}{-1} \right) - \int (1) \left( \frac{e^{-y}}{-1} \right) dy + c \) \( = -y e^{-y} + \int e^{-y} dy + c \)
Therefore, \( x e^{-y} = -y e^{-y} + \frac{e^{-y}}{-1} + c \)
Therefore, \( e^{-y} x = -y e^{-y} - e^{-y} + c \)
Therefore, \( e^{-y} x + y e^{-y} + e^{-y} = c \)
Therefore, \( e^{-y} (x + y + 1) = c \)
Therefore, \( x + y + 1 = c e^y \) This is the general solution.
In simple words: This differential equation is not directly in the \( \frac{dy}{dx} \) linear form. By taking the reciprocal, it transforms into a linear differential equation with respect to x and y, allowing the use of the integrating factor method. Integration by parts is necessary for solving the integral involving \( y e^{-y} \).

🎯 Exam Tip: If an equation is not linear in \( \frac{dy}{dx} \), consider if it can be made linear in \( \frac{dx}{dy} \) by inverting the derivative. This often simplifies the problem significantly. Be careful with the integration by parts formula, especially with negative signs.

Question 5. \( y dx + (x - y^2) dy = 0 \)
Answer: Solution: \( y dx + (x - y^2) dy = 0 \)
Therefore, \( y dx = -(x - y^2) dy \)
Therefore, \( \frac{dx}{dy} = - \frac{(x - y^2)}{y} = - \frac{x}{y} + y \)
Therefore, \( \frac{dx}{dy} + \left( \frac{1}{y} \right) \cdot x = y \)........(1) This is the linear differential equation of the form \( \frac{dx}{dy} + Px = Q \), where \( P = \frac{1}{y} \) and \( Q = y \)
Therefore, I.F. \( = e^{\int Pdy} = e^{\int \frac{1}{y} dy} = e^{\log y} = y \)
Therefore, the solution of (1) is given by \( x \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dy + c_1 \)
Therefore, \( xy = \int y \cdot y dy + c_1 \)
Therefore, \( xy = \int y^2 dy + c_1 \)
Therefore, \( xy = \frac{y^3}{3} + c_1 \)
Therefore, \( 3xy = y^3 + 3c_1 \)
Therefore, \( 3xy = y^3 + c \), where \( c = 3c_1 \) This is the general solution.
In simple words: This problem starts with a differential equation that can be rearranged into a linear form with respect to x and y. The integrating factor is found, and then the equation is solved by direct integration.

🎯 Exam Tip: Recognize when a differential equation is best solved by treating x as the dependent variable and y as the independent variable. This re-framing can simplify complex-looking equations into standard linear forms.

Question 6. \( \frac{dy}{dx} + 2xy = x \)
Answer: Solution: \( \frac{dy}{dx} + 2xy = x \).........(1) This is the linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = 2x, Q = x \)
Therefore, I.F. \( = e^{\int Pdx} = e^{\int 2x dx} \) \( = e^{2 \int x dx} = e^{2 \left( \frac{x^2}{2} \right)} = e^{x^2} \)
Therefore, the solution of (1) is given by \( y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dx + c \)
Therefore, \( y e^{x^2} = \int x e^{x^2} dx + c \) Put \( x^2 = t \)
Therefore, \( 2x dx = dt \)
Therefore, \( x dx = \frac{1}{2} dt \)
Therefore, (1) becomes \( y e^{x^2} = \int \frac{1}{2} e^t dt + c \)
Therefore, \( y e^{x^2} = \frac{1}{2} e^t + c \)
Therefore, \( y e^{x^2} = \frac{1}{2} e^{x^2} + c \) This is the general solution.
In simple words: This is a first-order linear differential equation solved using an integrating factor involving \( x^2 \). A substitution method is then applied to solve the resulting integral.

🎯 Exam Tip: Be mindful of substitutions when integrating. If the integral contains a composite function, look for its derivative to simplify the integration process. Double-check the back-substitution to express the final answer in terms of the original variable.

Question 7. \( (x + a) \frac{dy}{dx} = -y + a \)
Answer: Solution: \( (x + a) \frac{dy}{dx} = -y + a \)
Therefore, \( (x + a) \frac{dy}{dx} + y = a \)
Therefore, \( \frac{dy}{dx} + \left( \frac{1}{x+a} \right) y = \frac{a}{x+a} \)........(1) This is the linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{1}{x+a} \) and \( Q = \frac{a}{x+a} \)
Therefore, I.F. \( = e^{\int Pdx} = e^{\int \frac{1}{x+a} dx} \) \( = e^{\log (x+a)} = x + a \)
Therefore, the solution of (1) is given by \( y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dx + c \)
Therefore, \( y(x+a) = \int \left( \frac{a}{x+a} \right) (x+a) dx + c \) \( = a \int dx + c \)
Therefore, \( y(x+a) = ax + c \) This is the general solution.
In simple words: This linear differential equation needs to be rearranged into the standard form before finding the integrating factor. The integration step simplifies nicely as the integrating factor cancels out the denominator in the Q term.

🎯 Exam Tip: Always ensure the differential equation is in the standard linear form \( \frac{dy}{dx} + Py = Q \) before identifying P and Q. Be careful with algebraic manipulations to avoid errors in the integrating factor or the integral.

Question 8. \( dy + (2y) dx = 8 dx \)
Answer: Solution: \( dy + (2y) dx = 8 dx \)
Therefore, \( \frac{dy}{dx} + 2y = 8 \)........(1) This is the linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = 2, Q = 8 \)
Therefore, I.F. \( = e^{\int Pdx} = e^{\int 2 dx} = e^{2x} \)
Therefore, the solution of (1) is given by \( y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dx + c \)
Therefore, \( y e^{2x} = \int 8 e^{2x} dx + c \) \( = 8 \left( \frac{e^{2x}}{2} \right) + c \)
Therefore, \( y e^{2x} = 4 e^{2x} + c \) This is the general solution.
In simple words: This problem starts with an equation that needs rearrangement to form a linear differential equation. Once in standard form, the integrating factor is easily calculated, and the final solution is found through direct integration.

🎯 Exam Tip: Simple algebraic rearrangements are often the first crucial step in solving differential equations. Always group terms with dx and dy to isolate the derivative term and then manipulate it into the standard form. Ensure proper integration of exponential functions.