RBSE Solutions Class 12 Maths Chapter 12 अवकल समीकरण More Questions

 

Question 1. समीकरण \( (x^{2}+1) \frac { dy }{ dx } = 1 \) का हल है
(a) \( y = \cot^{-1}x + C \)
(b) \( y = \tan^{-1}x + C \)
(c) \( y = \sin^{-1}x + C \)
(d) \( y = \cos^{-1}x + C \)
Answer: (b) \( y = \tan^{-1}x + C \)
In simple words: First, separate the dy and dx terms. Then, integrate both sides of the equation. The integral of \( \frac { 1 }{ 1+x^{2} } \) is \( \tan^{-1}x \), which gives the final solution.

🎯 Exam Tip: Always remember the standard integration formulas, especially for inverse trigonometric functions, as they are frequently used in solving differential equations.

 

Question 2. समीकरण \( \frac { dy }{ dx } + 2x = e^{3x} \) का हल है
(a) \( y + x^{2} = \frac { 1 }{ 3 } e^{3x} + C \)
(b) \( y + x^{2} = \frac { 1 }{ 3 } e^{3x} + C \)
(c) \( y - x^{2} = \frac { 1 }{ 3 } e^{2x} + C \)
(d) \( y - x^{2} = \frac { 1 }{ 3 } e^{3x} + C \)
Answer: (a) \( y + x^{2} = \frac { 1 }{ 3 } e^{3x} + C \)
In simple words: Rearrange the equation to group terms with dy and dx. Then, integrate each part separately. The integral of \( e^{3x} \) is \( \frac { 1 }{ 3 } e^{3x} \), and the integral of \( 2x \) is \( x^{2} \). Add the constant C to get the general solution.

🎯 Exam Tip: When integrating exponential functions like \( e^{ax} \), remember the rule \( \int e^{ax} dx = \frac { 1 }{ a } e^{ax} \), and don't forget the constant of integration, C.

 

Question 3. समीकरण \( \frac { dy }{ dx } + \cos x \tan y = 0 \) का हल है
(a) \( \log \sin y + \sin x + C \)
Answer: (a) \( \log \sin y + \sin x + C \)
In simple words: First, move the term with \( \tan y \) and \( \cos x \) to the other side. Then, separate the variables by moving all y terms with dy and all x terms with dx. Integrate both sides: \( \int \cot y dy = \log | \sin y | \) and \( \int \cos x dx = \sin x \). Arrange the terms to form the final general solution.

🎯 Exam Tip: For variable separable differential equations, ensure all terms involving the dependent variable (y) are grouped with dy and all terms involving the independent variable (x) are grouped with dx before integrating.

 

Question 4. समीकरण \( \frac { dy }{ dx } = \frac { e^{x} + e^{-x} }{ e^{x} - e^{-x} } \) का हल है
(a) \( y = \log(e^{x} + e^{x}) + C \)
(b) \( y = \log (e^{x} - e^{-x}) + C \)
(c) \( y = \log (e^{x} + 1) + C \)
(d) \( y = \log (1 - e^{-x}) + C \)
Answer: (b) \( y = \log (e^{x} - e^{-x}) + C \)
In simple words: You can solve this by substituting the denominator as a new variable, say 't'. Then, the numerator becomes the differential of 't'. Integrating \( \frac { 1 }{ t } \) gives \( \log |t| \). Replace 't' back with the original expression to get the answer.

🎯 Exam Tip: Look for opportunities to use substitution (like \( t = \text{denominator} \) or \( t = \text{expression} \)) in integrals where the numerator is the derivative of the denominator, as this simplifies the integration to a logarithmic form.

 

Question 5. समीकरण \( e^{-x+y} \frac { dy }{ dx } = 1 \) का हल है
(a) \( e^{y} = e^{x} + C \)
(b) \( e^{y} = e^{-x} + C \)
(c) \( e^{-y} = e^{-x} + C \)
(d) \( e^{y} = e^{x} + C \)
Answer: (a) \( e^{y} = e^{x} + C \)
In simple words: First, rewrite \( e^{-x+y} \) as \( e^{y} e^{-x} \). Then, separate the variables so all terms with y are with dy and all terms with x are with dx. Integrate both \( e^{y} \) and \( e^{x} \) to get the final solution with a constant of integration.

🎯 Exam Tip: Always look to simplify exponential terms with sums or differences in the exponent by splitting them, like \( e^{a+b} = e^{a}e^{b} \) or \( e^{a-b} = e^{a}e^{-b} \), as this often helps in separating variables.

 

Question 6. समीकरण \( \frac { dy }{ dx } + \frac { 1 }{ x } y = 0 \) का हल है
(a) \( x + \frac { 1 }{ 2 } \log(1+y) + C \)
(b) \( x + \frac { 1 }{ 2 } \log(1+y^{2}) = C \)
(c) \( x + \log(1+y) = C \)
(d) \( x + \log(1+y^{2}) = C \)
Answer: (b) \( x + \frac { 1 }{ 2 } \log(1+y^{2}) = C \)
In simple words: The problem is solved by integrating both x and y terms separately. For the y term, you need to use a substitution method to integrate \( \frac { y }{ 1+y^{2} } \), which leads to a logarithm. Combine these integrals with a constant.

🎯 Exam Tip: When integrating fractions like \( \frac { y }{ 1+y^{2} } \), always consider a substitution where the derivative of the denominator (or part of it) is in the numerator, as this simplifies to \( \frac { 1 }{ 2 } \log|\text{denominator}| \).

 

Question 7. समीकरण \( \frac { dy }{ dx } = \cos^{2} y \) का हल है
(a) \( x + \tan y = C \)
(b) \( \tan y = x + C \)
(c) \( \sin y + x = C \)
(d) \( \sin y - x = C \)
Answer: (b) \( \tan y = x + C \)
In simple words: To solve this, separate the variables by moving \( \cos^{2} y \) to the left side with dy, making it \( \frac { 1 }{ \cos^{2} y } \), which is \( \sec^{2} y \). Then, integrate \( \sec^{2} y \) with respect to y, and 1 with respect to x. This will directly give the solution.

🎯 Exam Tip: Recognize common integral forms like \( \int \sec^{2} y dy = \tan y \). Knowing these standard integrals speeds up solving variable separable differential equations.

 

Question 8. समीकरण \( \frac { dy }{ dx } = e^{y} + x + e^{x} x^{2} \)
Answer: The problem implies solving the separable differential equation \( \frac { dy }{ dx } = e^{y}(e^{x} + x^{2}) \).
First, we separate the variables:
\( \frac { dy }{ e^{y} } = (e^{x} + x^{2}) dx \)
Next, we integrate both sides:
\( \int e^{-y} dy = \int (e^{x} + x^{2}) dx \)
Performing the integration:
\( -e^{-y} = e^{x} + \frac { x^{3} }{ 3 } + C \)
Rearranging the terms to match a common solution format:
\( e^{x} + e^{-y} + \frac { x^{3} }{ 3 } = C' \) (where \( C' = -C \))
In simple words: Start by rearranging the equation to separate the 'y' terms with dy and 'x' terms with dx. Then, integrate each side. Remember that integrating \( e^{-y} \) gives \( -e^{-y} \), and integrating \( e^{x} \) and \( x^{2} \) gives \( e^{x} \) and \( \frac { x^{3} }{ 3 } \) respectively. Don't forget the constant 'C'.

🎯 Exam Tip: When integrating \( e^{-y} \), treat it as \( \int e^{ay} dy \) where \( a=-1 \). The result is \( \frac { 1 }{ a } e^{ay} \). Always remember the sign changes when integrating with a negative exponent.

 

Question 9. अवकल समीकरण \( \frac { dy }{ dx } + \frac { y }{ x } = \frac { y^{2} }{ x^{2} } \) में निम्न में से किस प्रतिस्थापन द्वारा रैखिक समीकरण में परिवर्तित होगी ?
(a) \( y = t \)
(b) \( y^{2} = t \)
(c) \( \frac { 1 }{ y } = t \)
(d) \( \frac { 1 }{ y^{2} } = t \)
Answer: (c) \( \frac { 1 }{ y } = t \)
In simple words: This equation is a special type called Bernoulli's equation. To change it into a simpler linear form, you need to use a specific substitution. For this equation, setting \( t = \frac { 1 }{ y } \) will transform it into an easier form to solve.

🎯 Exam Tip: Identify Bernoulli's equations (form \( \frac { dy }{ dx } + P(x)y = Q(x)y^{n} \)) and remember the standard substitution \( v = y^{1-n} \) to convert them into linear differential equations.

 

Question 10. अवकल समीकरण \( \frac { dy }{ dx } + xy = e^{x} y^{3} \) में निम्न में से किस प्रतिस्थापन द्वारा अवकल समीकरण में परिवर्तित होगी
(a) \( \frac { 1 }{ y } = \upsilon \)

 

Question 11. अवकल समीकरण \( \frac { dy }{ dx } + 2x = e^{3x} \) का व्यापक हल ज्ञात कीजिए।
Answer: The given differential equation is \( \frac { dy }{ dx } + 2x = e^{3x} \).
First, we separate the variables by moving the \( 2x \) term to the right side:
\( \frac { dy }{ dx } = e^{3x} - 2x \)
Now, we multiply by dx to get all terms ready for integration:
\( dy = (e^{3x} - 2x) dx \)
Next, we integrate both sides:
\( \int dy = \int e^{3x} dx - \int 2x dx \)
Performing the integration:
\( y = \frac { e^{3x} }{ 3 } - \frac { 2x^{2} }{ 2 } + C \)
\( y = \frac { 1 }{ 3 } e^{3x} - x^{2} + C \)
We can rearrange this to:
\( y + x^{2} = \frac { 1 }{ 3 } e^{3x} + C \). This is the required general solution.In simple words: To find the general solution, first separate the 'dy' and 'dx' parts. Then, integrate each side of the equation. The integral of \( e^{3x} \) is \( \frac { e^{3x} }{ 3 } \), and the integral of \( 2x \) is \( x^{2} \). Add a constant C to the result to show all possible solutions.

🎯 Exam Tip: For problems asking for the 'general solution', remember to always include the constant of integration 'C'. This represents the family of all possible solutions to the differential equation.

 

Question 12. अवकल समीकरण \( \frac { dy }{ dx } + y \tan x = \sin x \) समाकलन गुणांक ज्ञात कीजिए।
Answer: The given differential equation is of the form \( \frac { dy }{ dx } + Py = Q \). Here, \( P = \tan x \). To find the integrating factor (I.F.), we use the formula \( I.F. = e^{\int P dx} \).
First, we find the integral of P:
\( \int P dx = \int \tan x dx = \log | \sec x | \)
Now, we calculate the integrating factor:
\( I.F. = e^{\log | \sec x |} \)
\( I.F. = \sec x \). The secant function is useful for solving these types of equations.In simple words: For the given equation, the 'P' part is \( \tan x \). We integrate \( \tan x \) to get \( \log | \sec x | \). Then, we put this into \( e^{\text{integral}} \), which simplifies to \( \sec x \). This value helps solve the full equation.

🎯 Exam Tip: Remember the standard form of a linear differential equation \( \frac { dy }{ dx } + Py = Q \) and the formula for the integrating factor, \( I.F. = e^{\int P dx} \). Correctly integrating P is crucial for full marks.

 

Question 13. अवकल समीकरण \( \frac { dy }{ dx } + \frac { 1 }{ \sin x } y = e^{x} \) का समाकलन गुणांक ज्ञात कीजिए।
Answer: The given differential equation is of the form \( \frac { dy }{ dx } + Py = Q \).
Here, \( P = \frac { 1 }{ \sin x } = \operatorname{cosec} x \) and \( Q = e^{x} \).
The integrating factor (समाकलन गुणांक) is given by \( I.F. = e^{\int P dx} \).
First, we find the integral of P:
\( \int P dx = \int \operatorname{cosec} x dx = \log | \operatorname{cosec} x - \cot x | \)
Now, substitute this into the I.F. formula:
\( I.F. = e^{\log | \operatorname{cosec} x - \cot x |} \)
\( I.F. = | \operatorname{cosec} x - \cot x | \)
This expression can be further simplified using trigonometric identities:
\( \operatorname{cosec} x - \cot x = \frac { 1 }{ \sin x } - \frac { \cos x }{ \sin x } = \frac { 1 - \cos x }{ \sin x } \)
Using half-angle formulas (\( 1 - \cos x = 2 \sin^{2} \frac { x }{ 2 } \) and \( \sin x = 2 \sin \frac { x }{ 2 } \cos \frac { x }{ 2 } \)):
\( = \frac { 2 \sin^{2} \frac { x }{ 2 } }{ 2 \sin \frac { x }{ 2 } \cos \frac { x }{ 2 } } = \frac { \sin \frac { x }{ 2 } }{ \cos \frac { x }{ 2 } } = \tan \frac { x }{ 2 } \). This simplification helps provide a more concise form.
Thus, \( I.F. = \tan \frac { x }{ 2 } \).In simple words: This is a linear differential equation. Identify 'P' as \( \operatorname{cosec} x \). Integrate 'P' to get \( \log | \operatorname{cosec} x - \cot x | \). Then, raise 'e' to this power. Simplify the resulting expression using trigonometric rules, which leads to \( \tan \frac { x }{ 2 } \).

🎯 Exam Tip: Always simplify the integrating factor to its most basic trigonometric form if possible. Remember the integral of \( \operatorname{cosec} x \) and the half-angle identity for \( \frac { 1 - \cos x }{ \sin x } \).

 

Question 14. अवकल समीकरण \( \cos(x+y) \frac { dy }{ dx } = 1 \) किस रूप की है ?
Answer: This differential equation is of the form that is reducible to variable separable form. By making a suitable substitution, the variables can be separated for integration. This makes the equation easier to solve in steps.
In simple words: This kind of equation can be changed into a form where you can easily separate the 'x' and 'y' parts. You need to use a simple change of variables to make it easy to solve.

🎯 Exam Tip: If an equation has terms like \( (ax+by+c) \), consider substituting the entire linear expression with a new variable to simplify it into a variable separable form.

 

Question 16. अवकल समीकरण \( \frac { dy }{ dx } = \frac { 4x+3y+1 }{ 3x+2y+1 } \) का व्यापक हल ज्ञात कीजिए।
Answer: The given differential equation is \( \frac { dy }{ dx } = \frac { 4x+3y+1 }{ 3x+2y+1 } \). This is a non-homogeneous equation reducible to homogeneous form because \( \frac { 4 }{ 3 } \neq \frac { 3 }{ 2 } \).
We use the substitution \( x = X+h \) and \( y = Y+k \). This implies \( dx = dX \) and \( dy = dY \).
Substituting these into the equation, we get:
\( \frac { dY }{ dX } = \frac { 4(X+h)+3(Y+k)+1 }{ 3(X+h)+2(Y+k)+1 } \)
\( \frac { dY }{ dX } = \frac { 4X+3Y+(4h+3k+1) }{ 3X+2Y+(3h+2k+1) } \)
To make the equation homogeneous, we set the constant terms to zero:
\( 4h+3k+1 = 0 \) ...(1)
\( 3h+2k+1 = 0 \) ...(2)
Solving these two linear equations for h and k:
Multiply equation (1) by 2: \( 8h+6k+2 = 0 \)
Multiply equation (2) by 3: \( 9h+6k+3 = 0 \)
Subtracting the first modified equation from the second: \( (9h-8h) + (6k-6k) + (3-2) = 0 \implies h+1 = 0 \implies h = -1 \).
Substitute \( h = -1 \) into equation (1): \( 4(-1)+3k+1 = 0 \implies -4+3k+1 = 0 \implies 3k-3 = 0 \implies 3k = 3 \implies k = 1 \).
So, \( h=-1 \) and \( k=1 \). The differential equation becomes:
\( \frac { dY }{ dX } = \frac { 4X+3Y }{ 3X+2Y } \). This is now a homogeneous differential equation.
For homogeneous equations, we substitute \( Y = vX \). Then, \( \frac { dY }{ dX } = v + X \frac { dv }{ dX } \).
Substituting this into the homogeneous equation:
\( v + X \frac { dv }{ dX } = \frac { 4X+3(vX) }{ 3X+2(vX) } \)
\( v + X \frac { dv }{ dX } = \frac { X(4+3v) }{ X(3+2v) } \)
\( X \frac { dv }{ dX } = \frac { 4+3v }{ 3+2v } - v \)
\( X \frac { dv }{ dX } = \frac { 4+3v - v(3+2v) }{ 3+2v } \)
\( X \frac { dv }{ dX } = \frac { 4+3v - 3v - 2v^{2} }{ 3+2v } \)
\( X \frac { dv }{ dX } = \frac { 4 - 2v^{2} }{ 3+2v } \)
Now, we separate the variables:
\( \frac { 3+2v }{ 4 - 2v^{2} } dv = \frac { dX }{ X } \)
Integrating both sides:
\( \int \frac { (3+2v) }{ 2(2 - v^{2}) } dv = \int \frac { dX }{ X } \)
The source's integration steps lead to:
\( \log (v^{2}-2) + \frac { 3 }{ 2\sqrt{2} } \log \left| \frac { v-\sqrt{2} }{ v+\sqrt{2} } \right| = -2 \log X + \log C \)
Substitute \( v = \frac { Y }{ X } \):
\( \log \left( \frac { Y^{2}-2X^{2} }{ X^{2} } \right) + \frac { 3 }{ 2\sqrt{2} } \log \left| \frac { Y-\sqrt{2}X }{ Y+\sqrt{2}X } \right| = \log \left( \frac { C }{ X^{2} } \right) \)
Combining the logarithms:
\( \log \left[ \frac { Y^{2}-2X^{2} }{ X^{2} } \cdot \left( \frac { Y-\sqrt{2}X }{ Y+\sqrt{2}X } \right)^{\frac { 3 }{ 2\sqrt{2} }} \right] = \log \left( \frac { C }{ X^{2} } \right) \)
This means:
\( (Y^{2}-2X^{2}) \left( \frac { Y-\sqrt{2}X }{ Y+\sqrt{2}X } \right)^{\frac { 3 }{ 2\sqrt{2} }} = C \)
Finally, substitute back \( Y = y-1 \) and \( X = x+1 \):
\( ((y-1)^{2}-2(x+1)^{2}) \left( \frac { (y-1)-\sqrt{2}(x+1) }{ (y-1)+\sqrt{2}(x+1) } \right)^{\frac { 3 }{ 2\sqrt{2} }} = C \). This is the general solution.In simple words: This problem involves changing a tricky equation into a simpler type. First, shift the origin by substituting x and y with new variables (X and Y) to make it a homogeneous equation. Then, use another substitution \( Y = vX \) to separate the variables (v and X). Integrate both sides, which is complex and involves logarithms. Finally, replace X and Y back with their original expressions involving x and y to get the answer.

🎯 Exam Tip: Equations of the form \( \frac { dy }{ dx } = \frac { ax+by+c }{ a'x+b'y+c' } \) are solved by substituting \( x=X+h \) and \( y=Y+k \) where h and k are found by setting \( ah+bk+c=0 \) and \( a'h+b'k+c'=0 \). This converts the equation to a homogeneous type.

 

Question 17. अवकल समीकरण \( \frac { dy }{ dx } = \frac { y }{ x } \left\{ \log \left( \frac { y }{ x } \right) + 1 \right\} \)
Answer: The given differential equation is \( \frac { dy }{ dx } = \frac { y }{ x } \left( \log \left( \frac { y }{ x } \right) + 1 \right) \). This is a homogeneous differential equation because it can be written as \( f(\frac { y }{ x }) \).
We make the substitution \( y = vx \). Then, by differentiating with respect to x, we get \( \frac { dy }{ dx } = v + x \frac { dv }{ dx } \).
Substitute \( y = vx \) and \( \frac { dy }{ dx } = v + x \frac { dv }{ dx } \) into the original equation:
\( v + x \frac { dv }{ dx } = v (\log v + 1) \)
\( v + x \frac { dv }{ dx } = v \log v + v \)
Subtracting v from both sides:
\( x \frac { dv }{ dx } = v \log v \)
Now, we separate the variables by moving v terms with dv and x terms with dx:
\( \frac { dv }{ v \log v } = \frac { dx }{ x } \)
Integrate both sides:
\( \int \frac { dv }{ v \log v } = \int \frac { dx }{ x } \)
For the left side, let \( t = \log v \). Then \( dt = \frac { 1 }{ v } dv \). So, \( \int \frac { dt }{ t } = \log |t| \).
Thus, the integration gives:
\( \log | \log v | = \log |x| + \log |C| \)
Using logarithm properties, \( \log | \log v | = \log |Cx| \)
Equating the arguments of the logarithm:
\( \log v = Cx \)
Finally, substitute back \( v = \frac { y }{ x } \):
\( \log \left( \frac { y }{ x } \right) = Cx \). This is the general solution.In simple words: This equation depends only on \( \frac { y }{ x } \). So, substitute \( y = vx \) and simplify. After simplifying, separate the 'v' and 'x' terms and integrate. Remember that \( \int \frac { 1 }{ x \log x } dx = \log | \log x | \). Then, put \( \frac { y }{ x } \) back in place of 'v' to get the final answer.

🎯 Exam Tip: For homogeneous differential equations (where \( \frac { dy }{ dx } = f(\frac { y }{ x }) \)), always use the substitution \( y = vx \). This transforms the equation into a variable separable form, making it solvable.

 

Question 18. \( \frac { dy }{ dx } = \frac { y + 2\sqrt{y^{2}-x^{2}} }{ x } \)
Answer: The given differential equation is \( \frac { dy }{ dx } = \frac { y + 2\sqrt{y^{2}-x^{2}} }{ x } \). This is a homogeneous differential equation as it can be expressed in terms of \( \frac { y }{ x } \).
We make the substitution \( y = vx \). By differentiating with respect to x, we get \( \frac { dy }{ dx } = v + x \frac { dv }{ dx } \).
Substitute \( y = vx \) and \( \frac { dy }{ dx } = v + x \frac { dv }{ dx } \) into the equation:
\( v + x \frac { dv }{ dx } = \frac { vx + 2\sqrt{(vx)^{2}-x^{2}} }{ x } \)
\( v + x \frac { dv }{ dx } = \frac { vx + 2\sqrt{x^{2}(v^{2}-1)} }{ x } \)
Taking x common in the numerator and canceling it with the denominator:
\( v + x \frac { dv }{ dx } = v + 2\sqrt{v^{2}-1} \)
Subtracting v from both sides:
\( x \frac { dv }{ dx } = 2\sqrt{v^{2}-1} \)
Now, we separate the variables:
\( \frac { dv }{ \sqrt{v^{2}-1} } = 2 \frac { dx }{ x } \)
Integrate both sides:
\( \int \frac { dv }{ \sqrt{v^{2}-1} } = \int 2 \frac { dx }{ x } \)
Using the standard integral formula \( \int \frac { 1 }{ \sqrt{x^{2}-a^{2}} } dx = \log |x + \sqrt{x^{2}-a^{2}}| \), with \( a=1 \):
\( \log |v + \sqrt{v^{2}-1}| = 2 \log |x| + \log |C| \)
Using logarithm properties, \( 2 \log |x| = \log |x^{2}| \):
\( \log |v + \sqrt{v^{2}-1}| = \log |Cx^{2}| \)
Equating the arguments of the logarithm:
\( v + \sqrt{v^{2}-1} = Cx^{2} \)
Finally, substitute back \( v = \frac { y }{ x } \):
\( \frac { y }{ x } + \sqrt{\left( \frac { y }{ x } \right)^{2}-1} = Cx^{2} \)
\( \frac { y }{ x } + \sqrt{\frac { y^{2}-x^{2} }{ x^{2} }} = Cx^{2} \)
Since \( \sqrt{x^{2}} = |x| \), we have \( \frac { y }{ x } + \frac { \sqrt{y^{2}-x^{2}} }{ x } = Cx^{2} \) (assuming \( x>0 \))
\( \frac { y + \sqrt{y^{2}-x^{2}} }{ x } = Cx^{2} \)
Multiplying by x:
\( y + \sqrt{y^{2}-x^{2}} = Cx^{3} \). This is the general solution.In simple words: This is a homogeneous equation, so start by substituting \( y = vx \) to simplify it. After separating the variables 'v' and 'x', integrate both sides. The integral of \( \frac { 1 }{ \sqrt{v^{2}-1} } \) is \( \log |v + \sqrt{v^{2}-1}| \). Finally, replace 'v' with \( \frac { y }{ x } \) and simplify the expression to get the answer.

🎯 Exam Tip: Remember the standard integral \( \int \frac { dx }{ \sqrt{x^{2}-a^{2}} } = \log |x + \sqrt{x^{2}-a^{2}}| \). Also, pay attention to the absolute value and the correct handling of \( \sqrt{x^{2}} = |x| \) during simplification.

 

Question 19. \( \frac { dy }{ dx } = e^{x-y} (e^{y} - e^{x}) \)
Answer: The given differential equation is \( \frac { dy }{ dx } = e^{x-y} (e^{y} - e^{x}) \).
First, simplify the exponential term: \( e^{x-y} = e^{x}e^{-y} \).
So, the equation becomes: \( \frac { dy }{ dx } = e^{x}e^{-y} (e^{y} - e^{x}) \)
Distribute \( e^{x}e^{-y} \) on the right side:
\( \frac { dy }{ dx } = e^{x}e^{-y}e^{y} - e^{x}e^{-y}e^{x} \)
\( \frac { dy }{ dx } = e^{x} - e^{2x}e^{-y} \)
Rearrange the terms to get a linear differential equation form \( \frac { dv }{ dx } + Pv = Q \). To do this, let \( v = e^{y} \).
Then, \( \frac { dv }{ dx } = e^{y} \frac { dy }{ dx } \). So, \( \frac { dy }{ dx } = \frac { 1 }{ e^{y}} \frac { dv }{ dx } = \frac { 1 }{ v } \frac { dv }{ dx } \).
Substitute into the rearranged equation, noting \( e^{-y} = \frac { 1 }{ v } \):
\( \frac { 1 }{ v } \frac { dv }{ dx } = e^{x} - e^{2x} \left( \frac { 1 }{ v } \right) \)
Multiply the entire equation by v:
\( \frac { dv }{ dx } = ve^{x} - e^{2x} \)
Rearrange into the standard linear form:
\( \frac { dv }{ dx } - e^{x}v = -e^{2x} \)
Here, \( P = -e^{x} \) and \( Q = -e^{2x} \).
The integrating factor (I.F.) is \( e^{\int P dx} \):
\( I.F. = e^{\int -e^{x} dx} = e^{-e^{x}} \)
The general solution for a linear equation is \( v \cdot (I.F.) = \int Q \cdot (I.F.) dx + C \).
\( v e^{-e^{x}} = \int (-e^{2x}) e^{-e^{x}} dx \)
To evaluate the integral, let \( t = e^{x} \). Then \( dt = e^{x} dx \). So, \( e^{2x} dx = e^{x} \cdot e^{x} dx = t dt \).
\( v e^{-e^{x}} = \int (-t) e^{-t} dt \)
Using integration by parts \( \int u dv = uv - \int v du \), with \( u=-t \) and \( dv=e^{-t} dt \):
\( \int (-t) e^{-t} dt = (-t)(-e^{-t}) - \int (-e^{-t})(-1) dt = te^{-t} - \int e^{-t} dt = te^{-t} - (-e^{-t}) = te^{-t} + e^{-t} \)
So, \( v e^{-e^{x}} = e^{-e^{x}}(e^{x}+1) + C \)
Divide by \( e^{-e^{x}} \) to solve for v:
\( v = e^{x} + 1 + C e^{e^{x}} \)
Finally, substitute back \( v = e^{y} \):
\( e^{y} = e^{x} + 1 + C e^{e^{x}} \). This is the general solution.In simple words: First, break down the exponential terms and rearrange the equation. Make a substitution \( v = e^{y} \) to turn it into a linear differential equation. Find the integrating factor for this new equation. Then, use the formula \( v \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dx + C \) to solve it. Finally, substitute \( e^{y} \) back for 'v' to get the answer.

🎯 Exam Tip: When a differential equation involves expressions like \( e^{y} \) or \( e^{-y} \) in a complex way, a substitution like \( v = e^{y} \) often simplifies it into a linear differential equation, which is easier to solve. Also, master integration by parts for integrals like \( \int t e^{t} dt \).

 

Question 20. \( \frac { dy }{ dx } + x \sin 2y = x^{3} \cos^{2} y \)
Answer: The given differential equation is \( \frac { dy }{ dx } + x \sin 2y = x^{3} \cos^{2} y \). This is a Bernoulli's equation.
First, divide the entire equation by \( \cos^{2} y \):
\( \frac { 1 }{ \cos^{2} y } \frac { dy }{ dx } + x \frac { \sin 2y }{ \cos^{2} y } = x^{3} \)
\( \sec^{2} y \frac { dy }{ dx } + x \frac { 2 \sin y \cos y }{ \cos^{2} y } = x^{3} \) (using \( \sin 2y = 2 \sin y \cos y \))
\( \sec^{2} y \frac { dy }{ dx } + 2x \tan y = x^{3} \)
Now, we make the substitution \( v = \tan y \). Differentiating with respect to x:
\( \frac { dv }{ dx } = \sec^{2} y \frac { dy }{ dx } \)
Substitute \( v = \tan y \) and \( \frac { dv }{ dx } \) into the equation:
\( \frac { dv }{ dx } + 2xv = x^{3} \)
This is a linear differential equation in v of the form \( \frac { dv }{ dx } + Pv = Q \).
Here, \( P = 2x \) and \( Q = x^{3} \).
The integrating factor (I.F.) is \( e^{\int P dx} \):
\( I.F. = e^{\int 2x dx} = e^{x^{2}} \)
The general solution for a linear equation is \( v \cdot (I.F.) = \int Q \cdot (I.F.) dx + C \).
\( v e^{x^{2}} = \int x^{3} e^{x^{2}} dx \)
To evaluate the integral \( \int x^{3} e^{x^{2}} dx \), use substitution. Let \( t = x^{2} \). Then \( dt = 2x dx \implies x dx = \frac { 1 }{ 2 } dt \).
The integral becomes \( \int x^{2} e^{x^{2}} (x dx) = \int t e^{t} \frac { 1 }{ 2 } dt = \frac { 1 }{ 2 } \int t e^{t} dt \).
Using integration by parts (\( \int u dv = uv - \int v du \)), with \( u=t \) and \( dv=e^{t} dt \):
\( \int t e^{t} dt = t e^{t} - \int e^{t} dt = t e^{t} - e^{t} = e^{t}(t-1) \)
So, the integral is \( \frac { 1 }{ 2 } e^{t}(t-1) \). Substitute back \( t = x^{2} \):
\( \frac { 1 }{ 2 } e^{x^{2}}(x^{2}-1) \)
Now, substitute this back into the general solution equation:
\( v e^{x^{2}} = \frac { 1 }{ 2 } e^{x^{2}}(x^{2}-1) + C \)
Finally, substitute back \( v = \tan y \):
\( \tan y e^{x^{2}} = \frac { 1 }{ 2 } e^{x^{2}}(x^{2}-1) + C \). This is the general solution.In simple words: This is a Bernoulli's equation. First, divide by \( \cos^{2} y \) and use the double angle formula for sine to change it into a linear equation. Then, substitute \( v = \tan y \). Solve this new linear equation by finding its integrating factor and integrating. Finally, replace 'v' with \( \tan y \) to get the solution.

🎯 Exam Tip: Recognize Bernoulli's equation and remember the transformation steps: divide by \( y^{n} \), then substitute \( v = y^{1-n} \). Also, be proficient with trigonometric identities like \( \sin 2y = 2 \sin y \cos y \) for simplification.

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