Samacheer Kalvi Class 12 Business Maths Solutions Chapter 4 Differential Equations Exercise 4.4

Question 1. Solve the differential equation: \( \frac { dy }{dx} + \left( \frac { -1 }{x} \right) y = x \)
Answer: The given differential equation is \( \frac { dy }{dx} + \left( \frac { -1 }{x} \right) y = x \). This equation is in the standard linear form \( \frac { dy }{dx} + Py = Q \).
Comparing, we have \( P = \frac { -1 }{x} \) and \( Q = x \).
First, we find the integrating factor (I.F.):
\( \int P dx = \int \frac { -1 }{x} dx = - \log x = \log \left( \frac { 1 }{x} \right) \)
\( I.F. = e^{\int P dx} = e^{\log \left( \frac { 1 }{x} \right)} = \frac { 1 }{x} \)
The general solution is given by \( y (I.F.) = \int Q (I.F.) dx + c \).
Substituting the values:
\( y \left( \frac { 1 }{x} \right) = \int x \left( \frac { 1 }{x} \right) dx + c \)
\( \implies y \left( \frac { 1 }{x} \right) = \int 1 dx + c \)
\( \implies \frac { y }{x} = x + c \)
So, the final solution is \( y = x^2 + cx \). Linear differential equations are crucial for modeling many real-world phenomena, from population growth to electrical circuits.
In simple words: We changed the equation into a standard form. Then, we found a special multiplying factor and used it to solve the equation. The answer shows how 'y' is related to 'x' and a constant 'c'.

🎯 Exam Tip: Always identify P and Q correctly after converting the given equation to the standard linear form \( \frac { dy }{dx} + Py = Q \). Be careful with the signs and logarithmic properties while calculating the Integrating Factor.

Question 2. Solve the differential equation: \( \frac { dy }{dx} + y \cos x = \sin x \cos x \)
Answer: The given differential equation is \( \frac { dy }{dx} + y \cos x = \sin x \cos x \). This is in the standard linear form \( \frac { dy }{dx} + Py = Q \).
Comparing, we have \( P = \cos x \) and \( Q = \sin x \cos x \).
First, we find the integrating factor (I.F.):
\( \int P dx = \int \cos x dx = \sin x \)
\( I.F. = e^{\int P dx} = e^{\sin x} \)
The general solution is given by \( y (I.F.) = \int Q (I.F.) dx + c \).
Substituting the values:
\( y (e^{\sin x}) = \int \sin x \cos x (e^{\sin x}) dx + c \)
To solve the integral \( \int \sin x \cos x (e^{\sin x}) dx \), let \( t = \sin x \). Then \( dt = \cos x dx \).
The integral becomes \( \int t e^t dt \).
Using integration by parts \( \int u dv = uv - \int v du \), let \( u = t \) and \( dv = e^t dt \).
Then \( du = dt \) and \( v = e^t \).
So, \( \int t e^t dt = t e^t - \int e^t dt = t e^t - e^t = e^t (t-1) \).
Substitute back \( t = \sin x \): \( e^{\sin x} (\sin x - 1) \).
Therefore, the solution is:
\( y (e^{\sin x}) = e^{\sin x} (\sin x - 1) + c \)
\( \implies y = (\sin x - 1) + c e^{-\sin x} \).
This type of problem demonstrates how a change of variable can simplify complex integrals.
In simple words: We identified the parts of the equation, found a special multiplying factor, and then used it to solve for 'y'. The main step was using a method called 'integration by parts' to solve a tricky integral.

🎯 Exam Tip: When solving \( \int Q (I.F.) dx \), always check if a substitution or integration by parts is required. Be careful with trigonometric derivatives and integrals.

Question 3. Solve the differential equation: \( x\frac { dy }{dx} + 2y = x^4 \)
Answer: The given differential equation is \( x\frac { dy }{dx} + 2y = x^4 \).
First, divide by 'x' to bring it to the standard linear form \( \frac { dy }{dx} + Py = Q \):
\( \frac { dy }{dx} + \frac { 2 }{x}y = x^3 \)
Comparing, we have \( P = \frac { 2 }{x} \) and \( Q = x^3 \).
Now, we find the integrating factor (I.F.):
\( \int P dx = \int \frac { 2 }{x} dx = 2 \int \frac { 1 }{x} dx = 2 \log x = \log x^2 \)
\( I.F. = e^{\int P dx} = e^{\log x^2} = x^2 \)
The general solution is given by \( y (I.F.) = \int Q (I.F.) dx + c \).
Substituting the values:
\( y (x^2) = \int x^3 (x^2) dx + c \)
\( \implies x^2 y = \int x^5 dx + c \)
\( \implies x^2 y = \frac { x^6 }{6} + c \)
This process is fundamental to solving many real-world problems involving rates of change, especially when they depend on both time and the quantity itself.
In simple words: We changed the given equation into a standard form by dividing by 'x'. Then we found a special multiplying factor (I.F.). Using this factor, we solved the equation to find how 'y' relates to 'x' and a constant.

🎯 Exam Tip: Always make sure the coefficient of \( \frac { dy }{dx} \) is 1 before identifying P and Q. Mistakes in this initial step lead to incorrect I.F. and solution.

Question 4. Solve the differential equation: \( \frac { dy }{dx} + \frac { 3x^2 }{1+x^3}y = \frac { 1+x^2 }{1+x^3} \)
Answer: The given differential equation is \( \frac { dy }{dx} + \frac { 3x^2 }{1+x^3}y = \frac { 1+x^2 }{1+x^3} \). This is already in the standard linear form \( \frac { dy }{dx} + Py = Q \).
Comparing, we have \( P = \frac { 3x^2 }{1+x^3} \) and \( Q = \frac { 1+x^2 }{1+x^3} \).
First, we find the integrating factor (I.F.):
\( \int P dx = \int \frac { 3x^2 }{1+x^3} dx \)
Let \( t = 1+x^3 \). Then \( dt = 3x^2 dx \).
So, \( \int \frac { 3x^2 }{1+x^3} dx = \int \frac { dt }{t} = \log t = \log (1+x^3) \).
\( I.F. = e^{\int P dx} = e^{\log (1+x^3)} = 1+x^3 \).
The general solution is given by \( y (I.F.) = \int Q (I.F.) dx + c \).
Substituting the values:
\( y (1+x^3) = \int \frac { 1+x^2 }{1+x^3} (1+x^3) dx + c \)
\( \implies y (1+x^3) = \int (1+x^2) dx + c \)
\( \implies y (1+x^3) = x + \frac { x^3 }{3} + c \)
This method is particularly useful when P and Q are functions of x, allowing for a systematic approach.
In simple words: We identified the 'P' and 'Q' parts of the equation, which was already in the correct form. We then calculated a special multiplying factor using the 'P' part. Finally, we solved the equation by integrating and found how 'y' depends on 'x'.

🎯 Exam Tip: When \( \int P dx \) involves a function and its derivative, use substitution for quicker integration. This problem uses \( 3x^2 \) as the derivative of \( 1+x^3 \).

Question 5. Solve the differential equation: \( \frac { dy }{dx} + \frac { y }{x} = x e^x \)
Answer: The given differential equation is \( \frac { dy }{dx} + \frac { 1 }{x}y = x e^x \). This is in the standard linear form \( \frac { dy }{dx} + Py = Q \).
Comparing, we have \( P = \frac { 1 }{x} \) and \( Q = x e^x \).
First, we find the integrating factor (I.F.):
\( \int P dx = \int \frac { 1 }{x} dx = \log x \)
\( I.F. = e^{\int P dx} = e^{\log x} = x \)
The general solution is given by \( y (I.F.) = \int Q (I.F.) dx + c \).
Substituting the values:
\( y(x) = \int (x e^x) (x) dx + c \)
\( \implies xy = \int x^2 e^x dx + c \)
To solve \( \int x^2 e^x dx \), we use integration by parts repeatedly.
Formula for repeated integration by parts: \( \int u v dx = uv - u'v_1 + u''v_2 - ... \)
Let \( u = x^2 \), so \( u' = 2x \), \( u'' = 2 \), \( u''' = 0 \).
Let \( dv = e^x dx \), so \( v = e^x \), \( v_1 = e^x \), \( v_2 = e^x \).
So, \( \int x^2 e^x dx = x^2 e^x - (2x) e^x + (2) e^x \)
\( = e^x (x^2 - 2x + 2) \).
Therefore, the solution is:
\( xy = e^x (x^2 - 2x + 2) + c \)
This method of solving linear differential equations is a cornerstone of mathematical physics and engineering.
In simple words: We put the equation in its standard form, found a special multiplying factor, and then integrated to find the solution. We used a special method called integration by parts twice to solve the integral involving \( x^2 e^x \).

🎯 Exam Tip: When integrating \( x^n e^x \) or \( x^n \sin x \), be prepared for repeated application of integration by parts. Tabular integration can be a faster alternative.

Question 6. Solve the differential equation: \( \frac { dy }{dx} + y \tan x = \cos^3 x \)
Answer: The given differential equation is \( \frac { dy }{dx} + y \tan x = \cos^3 x \). This is in the standard linear form \( \frac { dy }{dx} + Py = Q \).
Comparing, we have \( P = \tan x \) and \( Q = \cos^3 x \).
First, we find the integrating factor (I.F.):
\( \int P dx = \int \tan x dx = \int \frac { \sin x }{ \cos x } dx \)
Let \( t = \cos x \). Then \( dt = - \sin x dx \). So \( -dt = \sin x dx \).
\( \int \frac { -dt }{t} = - \log |t| = - \log |\cos x| = \log \left| \frac { 1 }{ \cos x } \right| = \log |\sec x| \).
\( I.F. = e^{\int P dx} = e^{\log |\sec x|} = \sec x \). (Assuming \( \cos x > 0 \)).
The general solution is given by \( y (I.F.) = \int Q (I.F.) dx + c \).
Substituting the values:
\( y (\sec x) = \int \cos^3 x (\sec x) dx + c \)
\( \implies y (\sec x) = \int \cos^3 x \left( \frac { 1 }{ \cos x } \right) dx + c \)
\( \implies y (\sec x) = \int \cos^2 x dx + c \)
We know that \( \cos^2 x = \frac { 1 + \cos 2x }{ 2 } \).
\( \implies y (\sec x) = \int \frac { 1 + \cos 2x }{ 2 } dx + c \)
\( \implies y (\sec x) = \frac { 1 }{2} \int (1 + \cos 2x) dx + c \)
\( \implies y (\sec x) = \frac { 1 }{2} \left( x + \frac { \sin 2x }{ 2 } \right) + c \)
This showcases how trigonometric identities are vital in simplifying integrals within differential equations.
In simple words: We identified P and Q from the equation. Then we calculated the special multiplying factor (I.F.) which turned out to be \( \sec x \). Finally, we integrated the right side, using a trigonometric identity for \( \cos^2 x \), to find the final relationship for 'y'.

🎯 Exam Tip: Remember key trigonometric identities like \( \cos^2 x = \frac { 1 + \cos 2x }{ 2 } \) and \( \sin^2 x = \frac { 1 - \cos 2x }{ 2 } \) as they are frequently used to simplify integrals involving powers of sine and cosine.

Question 7. If \( \frac { dy }{dx} + 2y \tan x = \sin x \) and if \( y = 0 \) when \( x = \pi/3 \) express \( y \) in terms of \( x \)
Answer: The given differential equation is \( \frac { dy }{dx} + 2y \tan x = \sin x \). This is in the standard linear form \( \frac { dy }{dx} + Py = Q \).
Comparing, we have \( P = 2 \tan x \) and \( Q = \sin x \).
First, we find the integrating factor (I.F.):
\( \int P dx = \int 2 \tan x dx = 2 \int \tan x dx = 2 \log |\sec x| = \log (\sec^2 x) \).
\( I.F. = e^{\int P dx} = e^{\log (\sec^2 x)} = \sec^2 x \).
The general solution is given by \( y (I.F.) = \int Q (I.F.) dx + c \).
Substituting the values:
\( y (\sec^2 x) = \int \sin x (\sec^2 x) dx + c \)
\( \implies y (\sec^2 x) = \int \sin x \left( \frac { 1 }{ \cos^2 x } \right) dx + c \)
\( \implies y (\sec^2 x) = \int \frac { \sin x }{ \cos x } \cdot \frac { 1 }{ \cos x } dx + c \)
\( \implies y (\sec^2 x) = \int \tan x \sec x dx + c \)
\( \implies y (\sec^2 x) = \sec x + c \) .......... (1)
Now, we use the initial condition: \( y = 0 \) when \( x = \pi/3 \).
Substitute these values into equation (1):
\( 0 \cdot \sec^2 (\pi/3) = \sec (\pi/3) + c \)
We know \( \sec (\pi/3) = \frac { 1 }{ \cos (\pi/3) } = \frac { 1 }{ 1/2 } = 2 \).
\( 0 \cdot (2)^2 = 2 + c \)
\( \implies 0 = 2 + c \)
\( \implies c = -2 \).
Substitute the value of 'c' back into equation (1):
\( y \sec^2 x = \sec x - 2 \)
\( \implies y = \frac { \sec x - 2 }{ \sec^2 x } \)
\( \implies y = \frac { \sec x }{ \sec^2 x } - \frac { 2 }{ \sec^2 x } \)
\( \implies y = \cos x - 2 \cos^2 x \).
Initial value problems are vital in physics for predicting future states based on current conditions.
In simple words: We solved the differential equation by finding its standard form, calculating the special multiplying factor, and then integrating. After getting a general solution with 'c', we used the given starting values (y=0 when x=π/3) to find the exact value of 'c'. This gave us the final answer for 'y' in terms of 'x'.

🎯 Exam Tip: For initial value problems, first find the general solution with the constant 'c'. Only then substitute the given initial conditions to find the specific value of 'c' and the particular solution.

Question 8. Solve the differential equation: \( \frac { dy }{dx} + \frac { y }{x} = x e^x \)
Answer: The given differential equation is \( \frac { dy }{dx} + \frac { 1 }{x}y = x e^x \). This is in the standard linear form \( \frac { dy }{dx} + Py = Q \).
Comparing, we have \( P = \frac { 1 }{x} \) and \( Q = x e^x \).
First, we find the integrating factor (I.F.):
\( \int P dx = \int \frac { 1 }{x} dx = \log x \)
\( I.F. = e^{\int P dx} = e^{\log x} = x \)
The general solution is given by \( y (I.F.) = \int Q (I.F.) dx + c \).
Substituting the values:
\( y(x) = \int (x e^x) (x) dx + c \)
\( \implies xy = \int x^2 e^x dx + c \)
To solve \( \int x^2 e^x dx \), we use integration by parts repeatedly.
Formula for repeated integration by parts: \( \int u v dx = uv - u'v_1 + u''v_2 - ... \)
Let \( u = x^2 \), so \( u' = 2x \), \( u'' = 2 \), \( u''' = 0 \).
Let \( dv = e^x dx \), so \( v = e^x \), \( v_1 = e^x \), \( v_2 = e^x \).
So, \( \int x^2 e^x dx = x^2 e^x - (2x) e^x + (2) e^x \)
\( = e^x (x^2 - 2x + 2) \).
Therefore, the solution is:
\( xy = e^x (x^2 - 2x + 2) + c \)
This problem highlights how the Integrating Factor method streamlines the solution of complex linear differential equations.
In simple words: We took the given equation, identified its parts, and calculated a special multiplying factor called the Integrating Factor. Then, we integrated both sides to find how 'y' relates to 'x' and a constant 'c'. The integration step involved using 'integration by parts' multiple times.

🎯 Exam Tip: Notice that this question is identical to Question 5. If encountered in an exam, ensure you solve it correctly and confidently, applying the same method.

Question 9. A bank pays interest by continuous compounding, that is by treating the interest rate as the instantaneous rate of change of principal. A man invests Rs 1,00,000 in the bank deposit which accrues interest, 8% over year compounded continuously. How much will he get after 10 years?
Answer: Let \( P(t) \) denote the amount of money in the account at time \( t \).
The differential equation governing continuous compounding at an 8% annual interest rate is:
\( \frac { dP }{dt} = \frac { 8 }{100} P \)
\( \implies \frac { dP }{dt} = 0.08 P \)
To solve this, we separate the variables:
\( \frac { dP }{P} = 0.08 dt \)
Now, we integrate both sides:
\( \int \frac { dP }{P} = \int 0.08 dt \)
\( \implies \log P = 0.08 t + c' \) (where \( c' \) is the integration constant)
To remove the logarithm, we use the exponential function:
\( P = e^{0.08 t + c'} \)
\( \implies P = e^{0.08 t} \cdot e^{c'} \)
Let \( C = e^{c'} \), which is also a constant.
So, \( P(t) = C e^{0.08 t} \) .......... (1)
We are given that initially (at \( t = 0 \)), the man invests Rs 1,00,000. So, \( P(0) = 1,00,000 \).
Substitute these values into equation (1):
\( 1,00,000 = C e^{0.08 \cdot 0} \)
\( \implies 1,00,000 = C e^0 \)
\( \implies 1,00,000 = C \cdot 1 \)
\( \implies C = 1,00,000 \).
Now, substitute the value of C back into equation (1):
\( P(t) = 1,00,000 e^{0.08 t} \)
We need to find the amount after 10 years, so we calculate \( P(10) \):
\( P(10) = 1,00,000 e^{0.08 \cdot 10} \)
\( \implies P(10) = 1,00,000 e^{0.8} \)
We are given that \( e^{0.8} = 2.2255 \).
\( P(10) = 1,00,000 \cdot 2.2255 \)
\( \implies P(10) = 2,22,550 \).
Therefore, the man will get Rs 2,22,550 after 10 years. Continuous compounding models are fundamental in finance for calculating exponential growth and decay.
In simple words: We used a special math equation to model how money grows when interest is added all the time. First, we solved this equation to get a general formula. Then, using the starting amount, we found the exact formula. Finally, we put in 10 years to find out how much money the man would have.

🎯 Exam Tip: Remember the formula for continuous compounding \( P(t) = P_0 e^{rt} \). For such problems, set up the differential equation, solve it, and then use the initial conditions to find the constant of integration.