Samacheer Kalvi Class 12 Maths Solutions Chapter 10 Ordinary Differential Equations Exercise 10.7

Question 1. Solve the following Linear differential equations.
\( \cos x \frac { dy }{ dx } + y \sin x = 1 \)
Answer: The given differential equation can be rewritten by dividing every term by \( \cos x \), which simplifies it into a standard linear form.
\( \cos x \frac { dy }{ dx } + y \sin x = 1 \)
Divide by \( \cos x \):
\( \frac { \cos x }{ \cos x } \frac { dy }{ dx } + \frac { y \sin x }{ \cos x } = \frac { 1 }{ \cos x } \)
\( \frac { dy }{ dx } + (\tan x) y = \sec x \)
This equation is now in the linear differential equation form: \( \frac { dy }{ dx } + Py = Q \)
Here, \( P = \tan x \) and \( Q = \sec x \).
Next, we find the integrating factor (I.F.):
\( I.F = e^{\int P dx} = e^{\int \tan x dx} \)
\( \implies I.F = e^{\log (\sec x)} \)
\( \implies I.F = \sec x \)
The general solution for a linear differential equation is given by:
\( y \times I.F = \int (Q \times I.F) dx + c \)
Substitute the values:
\( y \times \sec x = \int (\sec x \times \sec x) dx + c \)
\( y \sec x = \int \sec^2 x dx + c \)
\( y \sec x = \tan x + c \)
To find \( y \), we divide the entire equation by \( \sec x \):
\( y = \frac { \tan x }{ \sec x } + \frac { c }{ \sec x } \)
\( y = \frac { \sin x / \cos x }{ 1 / \cos x } + c \cos x \)
\( y = \sin x + c \cos x \)
This is the final solution for the given differential equation.
In simple words: We first change the equation into a special form. Then we find a multiplying factor that helps us solve it easily. Finally, we integrate and simplify to get the answer for \( y \).

🎯 Exam Tip: Remember to simplify the initial differential equation into the standard linear form \( \frac { dy }{ dx } + Py = Q \) (or \( \frac { dx }{ dy } + Px = Q \)) before calculating the integrating factor.

Question 2. \( (1 - x^2)\frac { dy }{ dx } - xy = 1 \)
Answer: The given differential equation is \( (1 - x^2)\frac { dy }{ dx } - xy = 1 \).
First, we rewrite the equation in the standard linear form \( \frac { dy }{ dx } + Py = Q \). To do this, divide all terms by \( (1 - x^2) \):
\( \frac { (1 - x^2) }{ (1 - x^2) } \frac { dy }{ dx } - \frac { xy }{ (1 - x^2) } = \frac { 1 }{ (1 - x^2) } \)
\( \frac { dy }{ dx } - \frac { x }{ (1 - x^2) } y = \frac { 1 }{ (1 - x^2) } \)
Now, we can identify \( P \) and \( Q \):
\( P = - \frac { x }{ (1 - x^2) } \)
\( Q = \frac { 1 }{ (1 - x^2) } \)
Next, we calculate the integrating factor (I.F.):
\( I.F = e^{\int P dx} = e^{\int - \frac { x }{ (1 - x^2) } dx} \)
To solve the integral, let \( u = 1 - x^2 \), then \( du = -2x dx \), so \( -x dx = \frac { 1 }{ 2 } du \).
The integral becomes \( \int \frac { 1 }{ u } \left( \frac { 1 }{ 2 } \right) du = \frac { 1 }{ 2 } \int \frac { 1 }{ u } du = \frac { 1 }{ 2 } \log |u| \).
Substitute \( u \) back:
\( \int - \frac { x }{ (1 - x^2) } dx = \frac { 1 }{ 2 } \log |1 - x^2| = \log ( (1 - x^2)^{\frac { 1 }{ 2 }} ) \)
\( \implies I.F = e^{\log ( \sqrt{1 - x^2} )} \)
\( \implies I.F = \sqrt{1 - x^2} \)
The solution is given by \( y \times I.F = \int (Q \times I.F) dx + c \):
\( y \sqrt{1 - x^2} = \int \left( \frac { 1 }{ 1 - x^2 } \times \sqrt{1 - x^2} \right) dx + c \)
\( y \sqrt{1 - x^2} = \int \frac { 1 }{ \sqrt{1 - x^2} } dx + c \)
The integral of \( \frac { 1 }{ \sqrt{1 - x^2} } \) is \( \sin^{-1} x \).
\( y \sqrt{1 - x^2} = \sin^{-1} x + c \)
To solve for \( y \), divide by \( \sqrt{1 - x^2} \):
\( y = \frac { \sin^{-1} x }{ \sqrt{1 - x^2} } + \frac { c }{ \sqrt{1 - x^2} } \)
\( y = \frac { \sin^{-1} x }{ \sqrt{1 - x^2} } + c (1 - x^2)^{-\frac { 1 }{ 2 }} \)
This is the required solution for the differential equation.
In simple words: We first rearrange the equation to a standard form, then find a special factor that helps us solve it. After that, we perform an integration and some algebra to find what \( y \) equals.

🎯 Exam Tip: When calculating the integrating factor for expressions like \( \frac { x }{ 1 - x^2 } \), use a substitution method (like \( u = 1 - x^2 \)) to simplify the integral in the exponent.

Question 3. \( \frac { dy }{ dx } + \frac { y }{ x } = \sin x \)
Answer: The given differential equation is \( \frac { dy }{ dx } + \frac { y }{ x } = \sin x \).
This equation is already in the standard linear form \( \frac { dy }{ dx } + Py = Q \).
Here, we can identify \( P \) and \( Q \):
\( P = \frac { 1 }{ x } \)
\( Q = \sin x \)
Next, we calculate the integrating factor (I.F.):
\( I.F = e^{\int P dx} = e^{\int \frac { 1 }{ x } dx} \)
\( \implies I.F = e^{\log x} \)
\( \implies I.F = x \)
The general solution for a linear differential equation is given by:
\( y \times I.F = \int (Q \times I.F) dx + c \)
Substitute the values:
\( yx = \int (\sin x \times x) dx + c \)
\( yx = \int x \sin x dx + c \)
To solve \( \int x \sin x dx \), we use integration by parts, which is a method to integrate products of functions. We choose \( u = x \) and \( dv = \sin x dx \). This means \( du = dx \) and \( v = -\cos x \).
Using the formula \( \int u dv = uv - \int v du \):
\( \int x \sin x dx = x(-\cos x) - \int (-\cos x) dx \)
\( \implies \int x \sin x dx = -x \cos x - (-\sin x) \)
\( \implies \int x \sin x dx = -x \cos x + \sin x \)
Now, substitute this back into the general solution equation:
\( yx = -x \cos x + \sin x + c \)
We can rearrange this to group terms with \( x \):
\( yx + x \cos x = \sin x + c \)
\( x(y + \cos x) = \sin x + c \)
This is the required solution for the differential equation.
In simple words: We saw the equation was already in the right shape. We found a special factor to multiply by. Then, we used a trick called "integration by parts" to solve the tricky part of the sum, and finally, we got the answer for \( y \).

🎯 Exam Tip: For integrals like \( \int x \sin x dx \), clearly show your steps for integration by parts by defining \( u, dv, du, \) and \( v \).

Question 4. \( (x^2 + 1)\frac { dy }{ dx } + 2xy = \sqrt {x^2+4 } \)
Answer: The given differential equation is \( (x^2 + 1)\frac { dy }{ dx } + 2xy = \sqrt {x^2+4 } \).
First, we need to convert it into the standard linear form \( \frac { dy }{ dx } + Py = Q \). We do this by dividing all terms by \( (x^2 + 1) \):
\( \frac { (x^2 + 1) }{ (x^2 + 1) } \frac { dy }{ dx } + \frac { 2xy }{ (x^2 + 1) } = \frac { \sqrt{x^2+4} }{ (x^2 + 1) } \)
\( \frac { dy }{ dx } + \frac { 2x }{ (x^2 + 1) } y = \frac { \sqrt{x^2+4} }{ (x^2 + 1) } \)
Now, we identify \( P \) and \( Q \):
\( P = \frac { 2x }{ (x^2 + 1) } \)
\( Q = \frac { \sqrt{x^2+4} }{ (x^2 + 1) } \)
Next, we compute the integrating factor (I.F.):
\( I.F = e^{\int P dx} = e^{\int \frac { 2x }{ (x^2 + 1) } dx} \)
For the integral \( \int \frac { 2x }{ (x^2 + 1) } dx \), let \( u = x^2 + 1 \), then \( du = 2x dx \). So, the integral becomes \( \int \frac { 1 }{ u } du = \log |u| \).
\( \int \frac { 2x }{ (x^2 + 1) } dx = \log (x^2 + 1) \)
\( \implies I.F = e^{\log (x^2 + 1)} \)
\( \implies I.F = x^2 + 1 \)
The general solution is given by \( y \times I.F = \int (Q \times I.F) dx + c \):
\( y (x^2 + 1) = \int \left( \frac { \sqrt{x^2+4} }{ (x^2 + 1) } \times (x^2 + 1) \right) dx + c \)
\( y (x^2 + 1) = \int \sqrt{x^2+4} dx + c \)
To solve \( \int \sqrt{x^2+4} dx \), we use the standard integral formula \( \int \sqrt{x^2+a^2} dx = \frac { x }{ 2 } \sqrt{x^2+a^2} + \frac { a^2 }{ 2 } \log |x + \sqrt{x^2+a^2}| \). Here, \( a^2 = 4 \).
\( y (x^2 + 1) = \frac { x }{ 2 } \sqrt{x^2+4} + \frac { 4 }{ 2 } \log |x + \sqrt{x^2+4}| + c \)
\( y (x^2 + 1) = \frac { x }{ 2 } \sqrt{x^2+4} + 2 \log |x + \sqrt{x^2+4}| + c \)
This is the required solution. This formula helps to directly integrate functions involving square roots of quadratic expressions.
In simple words: First, we change the equation into a standard form. Then we find a special factor to multiply by. After that, we use a known formula to integrate the square root part and then write down the final answer for \( y \).

🎯 Exam Tip: Memorize common integral formulas, especially those involving \( \sqrt{x^2 \pm a^2} \) or \( \sqrt{a^2 - x^2} \), as they frequently appear in solving differential equations.

Question 5. \( (2x – 10y^3) dy + y dx = 0 \)
Answer: The given differential equation is \( (2x – 10y^3) dy + y dx = 0 \).
This equation is not directly in the \( \frac { dy }{ dx } + Py = Q \) form. Let's try to rearrange it into the \( \frac { dx }{ dy } + Px = Q \) form, where \( x \) is the dependent variable.
First, move the \( y dx \) term:
\( (2x – 10y^3) dy = -y dx \)
Divide by \( dy \) to get \( \frac { dx }{ dy } \):
\( 2x – 10y^3 = -y \frac { dx }{ dy } \)
Rearrange the terms to get \( \frac { dx }{ dy } \) positive and \( x \) terms on one side:
\( y \frac { dx }{ dy } = -(2x – 10y^3) \)
\( y \frac { dx }{ dy } = -2x + 10y^3 \)
Divide by \( y \):
\( \frac { dx }{ dy } = - \frac { 2x }{ y } + \frac { 10y^3 }{ y } \)
\( \frac { dx }{ dy } = - \frac { 2 }{ y } x + 10y^2 \)
Move the \( x \) term to the left side:
\( \frac { dx }{ dy } + \frac { 2 }{ y } x = 10y^2 \)
This is now in the linear differential equation form \( \frac { dx }{ dy } + Px = Q \).
Here, \( P = \frac { 2 }{ y } \) and \( Q = 10y^2 \).
Next, we find the integrating factor (I.F.):
\( I.F = e^{\int P dy} = e^{\int \frac { 2 }{ y } dy} \)
\( \implies I.F = e^{2 \log y} \)
\( \implies I.F = e^{\log (y^2)} \)
\( \implies I.F = y^2 \)
The general solution for this form is given by \( x \times I.F = \int (Q \times I.F) dy + c \):
\( x y^2 = \int (10y^2 \times y^2) dy + c \)
\( x y^2 = \int 10y^4 dy + c \)
\( x y^2 = 10 \left( \frac { y^5 }{ 5 } \right) + c \)
\( x y^2 = 2y^5 + c \)
This is the required solution. Sometimes it's easier to switch the roles of dependent and independent variables.
In simple words: We changed the equation so \( x \) is the main variable we are solving for, not \( y \). Then we found a special factor and used it to find the final answer for \( x \) in terms of \( y \).

🎯 Exam Tip: If an equation isn't easily solvable in \( \frac { dy }{ dx } \) form, try rearranging it into \( \frac { dx }{ dy } \) form. This often works for equations that are linear in \( x \) but not in \( y \).

Question 6. \( x \sin x \frac { dy }{ dx } + (x \cos x + \sin x) y = \sin x \)
Answer: The given differential equation is \( x \sin x \frac { dy }{ dx } + (x \cos x + \sin x) y = \sin x \).
To transform this into the standard linear form \( \frac { dy }{ dx } + Py = Q \), we divide the entire equation by \( x \sin x \):
\( \frac { x \sin x }{ x \sin x } \frac { dy }{ dx } + \frac { (x \cos x + \sin x) }{ x \sin x } y = \frac { \sin x }{ x \sin x } \)
\( \frac { dy }{ dx } + \left( \frac { x \cos x }{ x \sin x } + \frac { \sin x }{ x \sin x } \right) y = \frac { 1 }{ x } \)
\( \frac { dy }{ dx } + \left( \cot x + \frac { 1 }{ x } \right) y = \frac { 1 }{ x } \)
Now, we identify \( P \) and \( Q \):
\( P = \cot x + \frac { 1 }{ x } \)
\( Q = \frac { 1 }{ x } \)
Next, we find the integrating factor (I.F.):
\( I.F = e^{\int P dx} = e^{\int (\cot x + \frac { 1 }{ x }) dx} \)
\( \implies I.F = e^{\int \cot x dx + \int \frac { 1 }{ x } dx} \)
\( \implies I.F = e^{\log (\sin x) + \log x} \)
Using logarithm properties, \( \log A + \log B = \log (AB) \):
\( \implies I.F = e^{\log (x \sin x)} \)
\( \implies I.F = x \sin x \)
The general solution is given by \( y \times I.F = \int (Q \times I.F) dx + c \):
\( y (x \sin x) = \int \left( \frac { 1 }{ x } \times x \sin x \right) dx + c \)
\( y (x \sin x) = \int \sin x dx + c \)
\( y (x \sin x) = -\cos x + c \)
We can rearrange this solution to isolate the terms involving \( x \sin x \):
\( xy \sin x + \cos x = c \)
This is the required solution. The integrating factor helps make the left side easy to integrate.
In simple words: We first put the equation into a standard form. Then we found a special multiplying factor. After that, we integrated both sides to find the answer for \( y \).

🎯 Exam Tip: Remember that \( \int \cot x dx = \log |\sin x| \) and \( \int \frac { 1 }{ x } dx = \log |x| \). Also, carefully apply logarithm properties like \( e^{\log f(x)} = f(x) \) and \( \log A + \log B = \log (AB) \).

Question 7. Solve \( (y - e^{\sin^{-1}x}) \frac { dx }{ dy } + \sqrt {1-x^2 } = 0 \)
Answer: The given differential equation is \( (y - e^{\sin^{-1}x}) \frac { dx }{ dy } + \sqrt {1-x^2 } = 0 \).
This equation is not in a standard linear form easily. Let's try to convert it into the \( \frac { dy }{ dx } + Py = Q \) form.
First, rearrange the terms:
\( (y - e^{\sin^{-1}x}) \frac { dx }{ dy } = - \sqrt{1-x^2} \)
Take the reciprocal of both sides to get \( \frac { dy }{ dx } \):
\( \frac { dy }{ dx } = \frac { -(y - e^{\sin^{-1}x}) }{ \sqrt{1-x^2} } \)
\( \frac { dy }{ dx } = \frac { e^{\sin^{-1}x} - y }{ \sqrt{1-x^2} } \)
Separate the terms on the right side:
\( \frac { dy }{ dx } = \frac { e^{\sin^{-1}x} }{ \sqrt{1-x^2} } - \frac { y }{ \sqrt{1-x^2} } \)
Move the term with \( y \) to the left side to get the standard linear form \( \frac { dy }{ dx } + Py = Q \):
\( \frac { dy }{ dx } + \frac { 1 }{ \sqrt{1-x^2} } y = \frac { e^{\sin^{-1}x} }{ \sqrt{1-x^2} } \)
Now, we identify \( P \) and \( Q \):
\( P = \frac { 1 }{ \sqrt{1-x^2} } \)
\( Q = \frac { e^{\sin^{-1}x} }{ \sqrt{1-x^2} } \)
Next, we find the integrating factor (I.F.):
\( I.F = e^{\int P dx} = e^{\int \frac { 1 }{ \sqrt{1-x^2} } dx} \)
\( \implies I.F = e^{\sin^{-1}x} \)
The general solution is given by \( y \times I.F = \int (Q \times I.F) dx + c \):
\( y e^{\sin^{-1}x} = \int \left( \frac { e^{\sin^{-1}x} }{ \sqrt{1-x^2} } \times e^{\sin^{-1}x} \right) dx + c \)
\( y e^{\sin^{-1}x} = \int \frac { e^{2 \sin^{-1}x} }{ \sqrt{1-x^2} } dx + c \)
To solve this integral, we use a substitution. Let \( t = \sin^{-1}x \). Then \( dt = \frac { 1 }{ \sqrt{1-x^2} } dx \).
The integral becomes \( \int e^{2t} dt \).
\( \int e^{2t} dt = \frac { e^{2t} }{ 2 } \)
Substitute \( t \) back:
\( \int \frac { e^{2 \sin^{-1}x} }{ \sqrt{1-x^2} } dx = \frac { e^{2 \sin^{-1}x} }{ 2 } \)
So, the solution is:
\( y e^{\sin^{-1}x} = \frac { e^{2 \sin^{-1}x} }{ 2 } + c \)
This is the required solution. Rearranging the original equation carefully is key to solving it.
In simple words: We first changed the equation's form to a standard one. Then we found a special multiplying factor. We also used a substitution trick for the integral part. Finally, we wrote the solution for \( y \).

🎯 Exam Tip: Be mindful of how \( \frac { dx }{ dy } \) relates to \( \frac { dy }{ dx } \). Often, reciprocating the differential term and rearranging can reveal a standard linear form that wasn't obvious initially.

Question 8. \( \frac { dy }{ dx } + \frac { y }{ (1-x)\sqrt{x} } = 1 - \sqrt{x} \)
Answer: The given differential equation is \( \frac { dy }{ dx } + \frac { y }{ (1-x)\sqrt{x} } = 1 - \sqrt{x} \).
This equation is already in the standard linear form \( \frac { dy }{ dx } + Py = Q \).
Here, we identify \( P \) and \( Q \):
\( P = \frac { 1 }{ (1-x)\sqrt{x} } \)
\( Q = 1 - \sqrt{x} \)
Next, we calculate the integrating factor (I.F.):
\( I.F = e^{\int P dx} = e^{\int \frac { 1 }{ (1-x)\sqrt{x} } dx} \)
To solve the integral \( \int \frac { 1 }{ (1-x)\sqrt{x} } dx \), we use a substitution. Let \( t = \sqrt{x} \). Then \( t^2 = x \), and \( 2t dt = dx \).
The integral becomes:
\( \int \frac { 1 }{ (1-t^2)t } (2t dt) = \int \frac { 2 }{ (1-t^2) } dt \)
\( \implies \int \frac { 2 }{ (1-t)(1+t) } dt \)
We use partial fractions for \( \frac { 2 }{ (1-t)(1+t) } = \frac { A }{ 1-t } + \frac { B }{ 1+t } \).
Multiplying by \( (1-t)(1+t) \), we get \( 2 = A(1+t) + B(1-t) \).
If \( t = 1 \), then \( 2 = A(2) \implies A = 1 \).
If \( t = -1 \), then \( 2 = B(2) \implies B = 1 \).
So, the integral is \( \int \left( \frac { 1 }{ 1-t } + \frac { 1 }{ 1+t } \right) dt \)
\( \implies - \log |1-t| + \log |1+t| \)
\( \implies \log \left| \frac { 1+t }{ 1-t } \right| \)
Substitute \( t = \sqrt{x} \) back:
\( \int P dx = \log \left| \frac { 1+\sqrt{x} }{ 1-\sqrt{x} } \right| \)
Therefore, the integrating factor is:
\( I.F = e^{\log \left| \frac { 1+\sqrt{x} }{ 1-\sqrt{x} } \right|} = \frac { 1+\sqrt{x} }{ 1-\sqrt{x} } \)
The general solution is given by \( y \times I.F = \int (Q \times I.F) dx + c \):
\( y \left( \frac { 1+\sqrt{x} }{ 1-\sqrt{x} } \right) = \int \left( (1-\sqrt{x}) \times \frac { 1+\sqrt{x} }{ 1-\sqrt{x} } \right) dx + c \)
\( y \left( \frac { 1+\sqrt{x} }{ 1-\sqrt{x} } \right) = \int (1+\sqrt{x}) dx + c \)
\( y \left( \frac { 1+\sqrt{x} }{ 1-\sqrt{x} } \right) = \int (1+x^{\frac { 1 }{ 2 }}) dx + c \)
\( y \left( \frac { 1+\sqrt{x} }{ 1-\sqrt{x} } \right) = x + \frac { x^{\frac { 1 }{ 2 } + 1} }{ \frac { 1 }{ 2 } + 1 } + c \)
\( y \left( \frac { 1+\sqrt{x} }{ 1-\sqrt{x} } \right) = x + \frac { x^{\frac { 3 }{ 2 }} }{ \frac { 3 }{ 2 } } + c \)
\( y \left( \frac { 1+\sqrt{x} }{ 1-\sqrt{x} } \right) = x + \frac { 2 }{ 3 } x^{\frac { 3 }{ 2 }} + c \)
Note that \( x^{\frac { 3 }{ 2 }} = x \sqrt{x} \).
\( y \left( \frac { 1+\sqrt{x} }{ 1-\sqrt{x} } \right) = x + \frac { 2 }{ 3 } x \sqrt{x} + c \)
This is the required solution. Substitution often simplifies complex integrals, making them solvable.
In simple words: We identified the equation's form and calculated a special multiplying factor using a substitution and partial fractions. Then we integrated to find the final answer for \( y \).

🎯 Exam Tip: For integrals involving \( \sqrt{x} \), try substituting \( t = \sqrt{x} \) (so \( x = t^2 \) and \( dx = 2t dt \)) to simplify the expression into a rational function, which can then be solved using partial fractions.

Question 9. \( (1 + x + xy^2) \frac { dy }{ dx } + (y + y^3) = 0 \)
Answer: The given differential equation is \( (1 + x + xy^2) \frac { dy }{ dx } + (y + y^3) = 0 \).
This equation is not linear in \( y \), but it can be made linear in \( x \). Let's rearrange it into the form \( \frac { dx }{ dy } + Px = Q \).
First, move the \( (y + y^3) \) term:
\( (1 + x + xy^2) \frac { dy }{ dx } = -(y + y^3) \)
Now, take the reciprocal of both sides to get \( \frac { dx }{ dy } \):
\( \frac { dx }{ dy } = \frac { -(1 + x + xy^2) }{ (y + y^3) } \)
Factor out common terms in the numerator and denominator:
\( \frac { dx }{ dy } = \frac { -(1 + x(1+y^2)) }{ y(1+y^2) } \)
Separate the terms on the right side:
\( \frac { dx }{ dy } = - \frac { 1 }{ y(1+y^2) } - \frac { x(1+y^2) }{ y(1+y^2) } \)
\( \frac { dx }{ dy } = - \frac { 1 }{ y(1+y^2) } - \frac { x }{ y } \)
Move the term with \( x \) to the left side:
\( \frac { dx }{ dy } + \frac { 1 }{ y } x = - \frac { 1 }{ y(1+y^2) } \)
This is now in the linear differential equation form \( \frac { dx }{ dy } + Px = Q \).
Here, \( P = \frac { 1 }{ y } \) and \( Q = - \frac { 1 }{ y(1+y^2) } \).
Next, we calculate the integrating factor (I.F.):
\( I.F = e^{\int P dy} = e^{\int \frac { 1 }{ y } dy} \)
\( \implies I.F = e^{\log y} \)
\( \implies I.F = y \)
The general solution for this form is given by \( x \times I.F = \int (Q \times I.F) dy + c \):
\( xy = \int \left( - \frac { 1 }{ y(1+y^2) } \times y \right) dy + c \)
\( xy = \int - \frac { 1 }{ 1+y^2 } dy + c \)
\( xy = - \int \frac { 1 }{ 1+y^2 } dy + c \)
The integral of \( \frac { 1 }{ 1+y^2} \) is \( \tan^{-1} y \).
\( xy = - \tan^{-1} y + c \)
Rearrange the solution:
\( xy + \tan^{-1} y = c \)
This is the required solution. Converting the equation to be linear in \( x \) was essential here.
In simple words: We changed the equation so we could solve for \( x \) instead of \( y \). Then we found a special multiplying factor, integrated, and simplified to get the answer.

🎯 Exam Tip: If an equation is not linear in \( y \), check if it's linear in \( x \) by rearranging it into the \( \frac { dx }{ dy } + Px = Q \) form. This is a common strategy for tricky differential equations.

Question 10. \( \frac { dy }{ dx } + \frac { y }{ x \log x } = \frac { \sin 2x }{ \log x } \)
Answer: The given differential equation is \( \frac { dy }{ dx } + \frac { y }{ x \log x } = \frac { \sin 2x }{ \log x } \).
This equation is already in the standard linear form \( \frac { dy }{ dx } + Py = Q \).
Here, we identify \( P \) and \( Q \):
\( P = \frac { 1 }{ x \log x } \)
\( Q = \frac { \sin 2x }{ \log x } \)
Next, we find the integrating factor (I.F.):
\( I.F = e^{\int P dx} = e^{\int \frac { 1 }{ x \log x } dx} \)
To solve the integral \( \int \frac { 1 }{ x \log x } dx \), we use substitution. Let \( t = \log x \). Then \( dt = \frac { 1 }{ x } dx \).
The integral becomes \( \int \frac { 1 }{ t } dt = \log |t| \).
Substitute \( t \) back:
\( \int \frac { 1 }{ x \log x } dx = \log |\log x| \)
\( \implies I.F = e^{\log |\log x|} \)
\( \implies I.F = \log x \)
The general solution is given by \( y \times I.F = \int (Q \times I.F) dx + c \):
\( y (\log x) = \int \left( \frac { \sin 2x }{ \log x } \times \log x \right) dx + c \)
\( y \log x = \int \sin 2x dx + c \)
The integral of \( \sin 2x \) is \( - \frac { \cos 2x }{ 2 } \).
\( y \log x = - \frac { \cos 2x }{ 2 } + c \)
We can rearrange this to:
\( y \log x + \frac { \cos 2x }{ 2 } = c \)
This is the required solution. Choosing the right substitution is crucial for simplifying the integration step.
In simple words: We recognized the equation's standard form and calculated a special multiplying factor using a substitution. After that, we integrated and arranged the terms to get the final answer.

🎯 Exam Tip: When \( P \) involves \( \log x \) or similar functions, a substitution for the integral \( \int P dx \) is often helpful. Also, remember the integral of \( \sin(ax) \) is \( -\frac{1}{a} \cos(ax) \).

Question 11. \( (x + a) \frac { dy }{ dx } – 2y = (x + a)^4 \)
Answer: The given differential equation is \( (x + a) \frac { dy }{ dx } – 2y = (x + a)^4 \).
To convert it into the standard linear form \( \frac { dy }{ dx } + Py = Q \), we divide the entire equation by \( (x + a) \):
\( \frac { (x + a) }{ (x + a) } \frac { dy }{ dx } - \frac { 2y }{ (x + a) } = \frac { (x + a)^4 }{ (x + a) } \)
\( \frac { dy }{ dx } - \frac { 2 }{ (x + a) } y = (x + a)^3 \)
Now, we identify \( P \) and \( Q \):
\( P = - \frac { 2 }{ (x + a) } \)
\( Q = (x + a)^3 \)
Next, we calculate the integrating factor (I.F.):
\( I.F = e^{\int P dx} = e^{\int - \frac { 2 }{ (x + a) } dx} \)
\( \implies I.F = e^{-2 \int \frac { 1 }{ (x + a) } dx} \)
\( \implies I.F = e^{-2 \log |x + a|} \)
\( \implies I.F = e^{\log ((x + a)^{-2})} \)
\( \implies I.F = (x + a)^{-2} = \frac { 1 }{ (x + a)^2 } \)
The general solution is given by \( y \times I.F = \int (Q \times I.F) dx + c \):
\( y \left( \frac { 1 }{ (x + a)^2 } \right) = \int \left( (x + a)^3 \times \frac { 1 }{ (x + a)^2 } \right) dx + c \)
\( \frac { y }{ (x + a)^2 } = \int (x + a) dx + c \)
\( \frac { y }{ (x + a)^2 } = \frac { (x + a)^2 }{ 2 } + c \)
To solve for \( y \), multiply both sides by \( (x + a)^2 \):
\( y = (x + a)^2 \left( \frac { (x + a)^2 }{ 2 } + c \right) \)
\( y = \frac { (x + a)^4 }{ 2 } + c (x + a)^2 \)
We can also write this as:
\( 2y = (x + a)^4 + 2c (x + a)^2 \)
\( 2y = (x + a)^2 [(x + a)^2 + 2c] \)
This is the required solution. Always ensure to correctly apply exponent rules when working with the integrating factor.
In simple words: We first put the equation into a standard form. Then we found a special multiplying factor using logarithm rules. After that, we integrated and rearranged the terms to get the final answer for \( y \).

🎯 Exam Tip: Be careful when dealing with constants in the exponent of the integrating factor, like the -2 in \( e^{-2 \log |x+a|} \). It becomes a power of the argument inside the logarithm, i.e., \( e^{\log((x+a)^{-2})} \).

Question 12. Solve the differential equation: \( \frac { dy }{ dx } = \frac { \sin^2 x }{ 1+x^3 } – \frac { 3x^2 }{ 1+x^3 }y \)
Answer: The given differential equation can be rearranged into the standard linear form \( \frac { dy }{ dx } + Py = Q \).
\( \frac { dy }{ dx } + \frac { 3x^2 }{ 1+x^3 }y = \frac { \sin^2 x }{ 1+x^3 } \)
Here, we identify \( P = \frac { 3x^2 }{ 1+x^3 } \) and \( Q = \frac { \sin^2 x }{ 1+x^3 } \).
Next, we calculate the Integrating Factor (I.F.) using the formula \( e^{\int P dx} \).
\( \int P dx = \int \frac { 3x^2 }{ 1+x^3 } dx \)
To solve this integral, we can use substitution. Let \( t = 1+x^3 \). Then, the derivative \( dt = 3x^2 dx \).
So, \( \int \frac { 1 }{ t } dt = \log |t| = \log |1+x^3| \).
Therefore, the Integrating Factor is \( I.F. = e^{\log |1+x^3|} = 1+x^3 \).
The general solution for a linear differential equation is given by \( y \times I.F = \int (Q \times I.F) dx + c \).
Substitute the values:
\( y (1+x^3) = \int \left( \frac { \sin^2 x }{ 1+x^3 } \right) (1+x^3) dx + c \)
\( y (1+x^3) = \int \sin^2 x dx + c \)
To integrate \( \sin^2 x \), we use the trigonometric identity \( \sin^2 x = \frac { 1 - \cos 2x }{ 2 } \).
\( y (1+x^3) = \int \frac { 1 - \cos 2x }{ 2 } dx + c \)
\( y (1+x^3) = \frac { 1 }{ 2 } \int (1 - \cos 2x) dx + c \)
\( y (1+x^3) = \frac { 1 }{ 2 } \left( x - \frac { \sin 2x }{ 2 } \right) + c \)
\( y (1+x^3) = \frac { x }{ 2 } - \frac { \sin 2x }{ 4 } + c \)
In simple words: First, we change the equation to a special standard form. Then, we find a helper term called the Integrating Factor. We use this helper term to solve the equation and find the final answer for y. This process makes complex equations easier to handle.

🎯 Exam Tip: Always remember to check for rearrangement into the standard linear form \( \frac { dy }{ dx } + Py = Q \) before finding the Integrating Factor.

Question 13. Solve the differential equation: \( x \frac { dy }{ dx } + y = x \log x \)
Answer: The given differential equation is \( x \frac { dy }{ dx } + y = x \log x \).
To convert it into the standard linear form \( \frac { dy }{ dx } + Py = Q \), we divide the entire equation by \( x \):
\( \frac { dy }{ dx } + \frac { 1 }{ x }y = \log x \)
Now, we identify \( P = \frac { 1 }{ x } \) and \( Q = \log x \).
Next, we calculate the Integrating Factor (I.F.) using the formula \( e^{\int P dx} \).
\( \int P dx = \int \frac { 1 }{ x } dx = \log |x| \).
Therefore, the Integrating Factor is \( I.F. = e^{\log |x|} = x \).
The general solution for a linear differential equation is given by \( y \times I.F = \int (Q \times I.F) dx + c \).
Substitute the values:
\( yx = \int (\log x \times x) dx + c \)
\( yx = \int x \log x dx + c \)
To integrate \( \int x \log x dx \), we use integration by parts, which is \( \int u dv = uv - \int v du \).
Let \( u = \log x \) and \( dv = x dx \).
Then, \( du = \frac { 1 }{ x } dx \) and \( v = \frac { x^2 }{ 2 } \).
\( \int x \log x dx = (\log x) \left( \frac { x^2 }{ 2 } \right) - \int \left( \frac { x^2 }{ 2 } \right) \left( \frac { 1 }{ x } \right) dx \)
\( = \frac { x^2 }{ 2 } \log x - \int \frac { x }{ 2 } dx \)
\( = \frac { x^2 }{ 2 } \log x - \frac { 1 }{ 2 } \left( \frac { x^2 }{ 2 } \right) + c \)
\( = \frac { x^2 }{ 2 } \log x - \frac { x^2 }{ 4 } + c \)
So, the solution is:
\( yx = \frac { x^2 }{ 2 } \log x - \frac { x^2 }{ 4 } + c \)
Multiplying the entire equation by 4 to remove fractions gives:
\( 4yx = 2x^2 \log x - x^2 + 4c \)
This can also be written as \( 4xy = 2x^2 \log x - x^2 + 4c \).
In simple words: We first make the equation fit a standard form. Then we find a special multiplier called the Integrating Factor. We use a method called integration by parts to solve a tough integral, and finally, we get the complete answer for y.

🎯 Exam Tip: Remember to simplify the equation to the standard form \( \frac { dy }{ dx } + Py = Q \) by dividing by the coefficient of \( \frac{dy}{dx} \) if it's not 1.

Question 14. Solve the differential equation: \( x \frac { dy }{ dx } + 2y – x² \log x = 0 \)
Answer: The given differential equation is \( x \frac { dy }{ dx } + 2y – x^2 \log x = 0 \).
First, we rearrange it into the standard linear form \( \frac { dy }{ dx } + Py = Q \).
\( x \frac { dy }{ dx } + 2y = x^2 \log x \)
Divide the entire equation by \( x \):
\( \frac { dy }{ dx } + \frac { 2 }{ x }y = x \log x \)
Now, we identify \( P = \frac { 2 }{ x } \) and \( Q = x \log x \).
Next, we calculate the Integrating Factor (I.F.) using the formula \( e^{\int P dx} \).
\( \int P dx = \int \frac { 2 }{ x } dx = 2 \log |x| = \log (x^2) \).
Therefore, the Integrating Factor is \( I.F. = e^{\log (x^2)} = x^2 \).
The general solution for a linear differential equation is given by \( y \times I.F = \int (Q \times I.F) dx + c \).
Substitute the values:
\( yx^2 = \int (x \log x \times x^2) dx + c \)
\( yx^2 = \int x^3 \log x dx + c \)
To integrate \( \int x^3 \log x dx \), we use integration by parts, which is \( \int u dv = uv - \int v du \).
Let \( u = \log x \) and \( dv = x^3 dx \).
Then, \( du = \frac { 1 }{ x } dx \) and \( v = \frac { x^4 }{ 4 } \).
\( \int x^3 \log x dx = (\log x) \left( \frac { x^4 }{ 4 } \right) - \int \left( \frac { x^4 }{ 4 } \right) \left( \frac { 1 }{ x } \right) dx \)
\( = \frac { x^4 }{ 4 } \log x - \int \frac { x^3 }{ 4 } dx \)
\( = \frac { x^4 }{ 4 } \log x - \frac { 1 }{ 4 } \left( \frac { x^4 }{ 4 } \right) + c \)
\( = \frac { x^4 }{ 4 } \log x - \frac { x^4 }{ 16 } + c \)
So, the solution is:
\( yx^2 = \frac { x^4 }{ 4 } \log x - \frac { x^4 }{ 16 } + c \)
Multiplying the entire equation by 16 to clear fractions gives:
\( 16yx^2 = 4x^4 \log x - x^4 + 16c \)
This can also be written as \( 16x^2y = 4x^4 \log x - x^4 + 16c \). This method is useful for solving a wide variety of first-order linear differential equations.
In simple words: We first rewrite the equation to a standard linear form. Then we find an Integrating Factor. We use a special method called integration by parts to solve the remaining integral. Finally, we simplify the answer to get the solution for y.

🎯 Exam Tip: When using integration by parts, correctly choosing \( u \) and \( dv \) is crucial. A common strategy is to choose \( u \) as the part that simplifies when differentiated (like \( \log x \)).

Question 15. Solve the differential equation: \( \frac { dy }{ dx } + \frac { 3y }{ x } = \frac { 1 }{ x^2 } \), given that \( y = 2 \) when \( x = 1 \)
Answer: The given differential equation is \( \frac { dy }{ dx } + \frac { 3y }{ x } = \frac { 1 }{ x^2 } \).
This equation is already in the standard linear form \( \frac { dy }{ dx } + Py = Q \).
Here, we identify \( P = \frac { 3 }{ x } \) and \( Q = \frac { 1 }{ x^2 } \).
Next, we calculate the Integrating Factor (I.F.) using the formula \( e^{\int P dx} \).
\( \int P dx = \int \frac { 3 }{ x } dx = 3 \log |x| = \log (x^3) \).
Therefore, the Integrating Factor is \( I.F. = e^{\log (x^3)} = x^3 \).
The general solution for a linear differential equation is given by \( y \times I.F = \int (Q \times I.F) dx + c \).
Substitute the values:
\( yx^3 = \int \left( \frac { 1 }{ x^2 } \times x^3 \right) dx + c \)
\( yx^3 = \int x dx + c \)
Integrate \( x \):
\( \int x dx = \frac { x^2 }{ 2 } + c \)
So, the general solution is:
\( yx^3 = \frac { x^2 }{ 2 } + c \)
Now, we use the given condition that \( y = 2 \) when \( x = 1 \) to find the value of \( c \).
Substitute \( x = 1 \) and \( y = 2 \) into the general solution:
\( (2)(1)^3 = \frac { (1)^2 }{ 2 } + c \)
\( 2 \times 1 = \frac { 1 }{ 2 } + c \)
\( 2 = \frac { 1 }{ 2 } + c \)
Subtract \( \frac { 1 }{ 2 } \) from both sides to find \( c \):
\( c = 2 - \frac { 1 }{ 2 } = \frac { 4 }{ 2 } - \frac { 1 }{ 2 } = \frac { 3 }{ 2 } \)
Now, substitute the value of \( c = \frac { 3 }{ 2 } \) back into the general solution:
\( yx^3 = \frac { x^2 }{ 2 } + \frac { 3 }{ 2 } \)
To clear the fractions, multiply the entire equation by 2:
\( 2yx^3 = x^2 + 3 \)
In simple words: We find a special multiplier called the Integrating Factor. We use it to solve the main part of the equation and find a general answer with 'c'. Then, we use the given numbers for x and y to find the exact value of 'c', which gives us the final, specific solution.

🎯 Exam Tip: For problems with initial conditions, always find the general solution first, then substitute the given values of \( x \) and \( y \) to determine the specific constant \( c \).