RBSE Solutions Class 12 Maths Chapter 12 Differential Equation Exercise 12.7

Solve the following differential equations:

Question 1. \( \frac { dy }{ dx } + \frac { 3x+2y-5 }{ 2x+3y-5 } = 0 \)
Answer: The given differential equation can be rewritten as:
\[ \frac { dy }{ dx } = - \frac { 3x+2y-5 }{ 2x+3y-5 } \] To solve this non-homogeneous equation, we substitute \( x = X+h \) and \( y = Y+k \).
This changes the derivative to \( \frac { dy }{ dx } = \frac { dY }{ dX } \).
So, \( \frac { dY }{ dX } = - \frac { 3(X+h)+2(Y+k)-5 }{ 2(X+h)+3(Y+k)-5 } \)
\[ \frac { dY }{ dX } = - \frac { 3X+2Y+(3h+2k-5) }{ 2X+3Y+(2h+3k-5) } \] For this equation to become homogeneous, the constant terms in the numerator and denominator must be zero:
\( 3h+2k-5 = 0 \)
\( 2h+3k-5 = 0 \) Solving these two linear equations simultaneously gives:
\( h = 1 \) and \( k = 1 \) Now the differential equation becomes homogeneous:
\[ \frac { dY }{ dX } = - \frac { 3X+2Y }{ 2X+3Y } \] We substitute \( Y = vX \), which means \( \frac { dY }{ dX } = v + X \frac { dv }{ dX } \).
\( v + X \frac { dv }{ dX } = - \frac { 3X+2vX }{ 2X+3vX } \)
\( v + X \frac { dv }{ dX } = - \frac { X(3+2v) }{ X(2+3v) } \)
\( v + X \frac { dv }{ dX } = - \frac { 3+2v }{ 2+3v } \)
\( X \frac { dv }{ dX } = - \frac { 3+2v }{ 2+3v } - v \)
\( X \frac { dv }{ dX } = \frac { -(3+2v) - v(2+3v) }{ 2+3v } \)
\( X \frac { dv }{ dX } = \frac { -3-2v-2v-3v^2 }{ 2+3v } \)
\( X \frac { dv }{ dX } = \frac { -3-4v-3v^2 }{ 2+3v } \) Now, separate the variables:
\[ \frac { dX }{ X } = \frac { (2+3v) }{ -(3v^2+4v+3) } dv \] Integrate both sides:
\[ \int \frac { dX }{ X } = \int - \frac { (2+3v) }{ (3v^2+4v+3) } dv \] The derivative of the denominator \( 3v^2+4v+3 \) is \( 6v+4 \). We can rewrite the numerator \( (2+3v) \) as \( \frac { 1 }{ 2 } (6v+4) \).
\[ \int \frac { dX }{ X } = - \frac { 1 }{ 2 } \int \frac { (6v+4) }{ (3v^2+4v+3) } dv \]
\( \log|X| = - \frac { 1 }{ 2 } \log|3v^2+4v+3| + \log c \) Multiply by 2:
\( 2 \log|X| = - \log|3v^2+4v+3| + 2 \log c \)
\( \log(X^2) = \log \left( \frac { c^2 }{ 3v^2+4v+3 } \right) \)
\( X^2 = \frac { C }{ 3v^2+4v+3 } \) (where \( C = c^2 \))
\( X^2 (3v^2+4v+3) = C \) Substitute \( v = \frac { Y }{ X } \):
\( X^2 \left( 3\left(\frac{Y}{X}\right)^2 + 4\left(\frac{Y}{X}\right) + 3 \right) = C \)
\( X^2 \left( \frac{3Y^2}{X^2} + \frac{4Y}{X} + 3 \right) = C \)
\( 3Y^2 + 4XY + 3X^2 = C \) Finally, substitute back \( X = x-1 \) and \( Y = y-1 \):
\( 3(y-1)^2 + 4(x-1)(y-1) + 3(x-1)^2 = C \) Expand this expression:
\( 3(y^2-2y+1) + 4(xy-x-y+1) + 3(x^2-2x+1) = C \)
\( 3y^2-6y+3 + 4xy-4x-4y+4 + 3x^2-6x+3 = C \)
\( 3x^2 + 3y^2 + 4xy - 10x - 10y + 10 = C \) This is the general solution for the given differential equation. The technique involves shifting the origin to eliminate constant terms, making the equation homogeneous.
In simple words: We first change the variables to make the equation simpler. Then we separate the parts with \(x\) and \(y\) and integrate both sides. This gives us the final equation which is the solution.

🎯 Exam Tip: Remember to solve the system of equations for 'h' and 'k' accurately, as any error here will propagate through the entire solution. Also, be careful with algebraic manipulations after substituting \( Y = vX \).

Question 2. Here \( \frac { dy }{ dx } = \frac { x-y+3 }{ 2(x-y)+5 } \)
Answer: The given differential equation is:
\[ \frac { dy }{ dx } = \frac { x-y+3 }{ 2(x-y)+5 } \] We notice that \( (x-y) \) appears in both the numerator and denominator. This suggests a substitution. Let \( v = x-y \). Then, differentiate both sides with respect to \( x \):
\( \frac { dv }{ dx } = 1 - \frac { dy }{ dx } \)
So, \( \frac { dy }{ dx } = 1 - \frac { dv }{ dx } \) Substitute this into the original differential equation:
\( 1 - \frac { dv }{ dx } = \frac { v+3 }{ 2v+5 } \) Now, we solve for \( \frac { dv }{ dx } \):
\( \frac { dv }{ dx } = 1 - \frac { v+3 }{ 2v+5 } \)
\( \frac { dv }{ dx } = \frac { (2v+5) - (v+3) }{ 2v+5 } \)
\( \frac { dv }{ dx } = \frac { 2v+5-v-3 }{ 2v+5 } \)
\( \frac { dv }{ dx } = \frac { v+2 }{ 2v+5 } \) Now, separate the variables:
\[ dx = \frac { 2v+5 }{ v+2 } dv \] We can perform polynomial long division or rewrite the numerator:
\( \frac { 2v+5 }{ v+2 } = \frac { 2(v+2)+1 }{ v+2 } = 2 + \frac { 1 }{ v+2 } \) So, the equation becomes:
\[ dx = \left( 2 + \frac { 1 }{ v+2 } \right) dv \] Integrate both sides:
\[ \int dx = \int \left( 2 + \frac { 1 }{ v+2 } \right) dv \]
\( x+C = 2v + \log|v+2| \) Finally, substitute back \( v = x-y \):
\( x+C = 2(x-y) + \log|x-y+2| \) Rearrange the terms to get the final solution:
\( x - 2(x-y) + \log|x-y+2| = C \)
\( x - 2x + 2y + \log|x-y+2| = C \)
\( -x + 2y + \log|x-y+2| = C \) This is the required solution. The substitution technique simplifies the equation, making it easier to integrate.
In simple words: We saw that 'x-y' was common, so we replaced it with 'v'. This made the equation simpler. After some algebra, we separated 'x' terms from 'v' terms and integrated them. Then we put 'x-y' back for 'v' to get the final answer.

🎯 Exam Tip: When you see repeated linear expressions like \( (ax+by) \) or \( (ax+by+c) \) in a differential equation, consider a substitution to simplify the problem, as it often reduces to a separable variable form.

Question 3. \( (2x + y + 1) dx + (4x + 2y - 1) dy = 0 \)
Answer: The given differential equation is:
\[ (2x + y + 1) dx + (4x + 2y - 1) dy = 0 \] First, express \( \frac { dy }{ dx } \):
\[ \frac { dy }{ dx } = - \frac { 2x+y+1 }{ 4x+2y-1 } \] Notice the expression \( (2x+y) \) appears in both the numerator and denominator. This indicates a suitable substitution. Let \( v = 2x+y \). Differentiate both sides with respect to \( x \):
\( \frac { dv }{ dx } = 2 + \frac { dy }{ dx } \) So, \( \frac { dy }{ dx } = \frac { dv }{ dx } - 2 \) Substitute this into the differential equation:
\[ \frac { dv }{ dx } - 2 = - \frac { v+1 }{ 2v-1 } \] Now, solve for \( \frac { dv }{ dx } \):
\[ \frac { dv }{ dx } = 2 - \frac { v+1 }{ 2v-1 } \]
\[ \frac { dv }{ dx } = \frac { 2(2v-1) - (v+1) }{ 2v-1 } \]
\[ \frac { dv }{ dx } = \frac { 4v-2-v-1 }{ 2v-1 } \]
\[ \frac { dv }{ dx } = \frac { 3v-3 }{ 2v-1 } \] Now, separate the variables:
\[ dx = \frac { 2v-1 }{ 3v-3 } dv \] We can factor out 3 from the denominator:
\[ dx = \frac { 1 }{ 3 } \frac { 2v-1 }{ v-1 } dv \] Rewrite the fraction \( \frac { 2v-1 }{ v-1 } \):
\( \frac { 2v-1 }{ v-1 } = \frac { 2(v-1)+1 }{ v-1 } = 2 + \frac { 1 }{ v-1 } \) So, the equation becomes:
\[ dx = \frac { 1 }{ 3 } \left( 2 + \frac { 1 }{ v-1 } \right) dv \] Integrate both sides:
\[ \int dx = \int \frac { 1 }{ 3 } \left( 2 + \frac { 1 }{ v-1 } \right) dv \]
\( x+C_1 = \frac { 1 }{ 3 } (2v + \log|v-1|) \) Multiply by 3:
\( 3x+3C_1 = 2v + \log|v-1| \) Let \( C = 3C_1 \) be a new arbitrary constant:
\( 3x+C = 2v + \log|v-1| \) Finally, substitute back \( v = 2x+y \):
\( 3x+C = 2(2x+y) + \log|2x+y-1| \)
\( 3x+C = 4x+2y + \log|2x+y-1| \) Rearrange the terms to get the final solution:
\( C = (4x-3x) + 2y + \log|2x+y-1| \)
\( C = x+2y + \log|2x+y-1| \) This is the required solution. The substitution makes the equation separable and integrable.
In simple words: We noticed '2x+y' was common, so we replaced it with 'v' to simplify. After reorganizing the terms, we could separate the 'x' and 'v' parts and integrate them. Then we put '2x+y' back for 'v' to get the final answer.

🎯 Exam Tip: Look for repeated linear combinations of variables (like \( ax+by \)) in non-homogeneous equations, as these are strong indicators for a direct substitution to simplify the problem.

Question 4. \( \frac { dy }{ dx } = \frac { 1-3x-3y }{ 2(x+y) } \)
Answer: The given differential equation is:
\[ \frac { dy }{ dx } = \frac { 1-3x-3y }{ 2(x+y) } \] Notice that \( (x+y) \) appears in both the numerator and the denominator. This suggests a substitution. Rewrite the numerator: \( 1-3x-3y = 1-3(x+y) \). So the equation becomes:
\[ \frac { dy }{ dx } = \frac { 1-3(x+y) }{ 2(x+y) } \] Let \( v = x+y \). Differentiate both sides with respect to \( x \):
\( \frac { dv }{ dx } = 1 + \frac { dy }{ dx } \) So, \( \frac { dy }{ dx } = \frac { dv }{ dx } - 1 \) Substitute this into the differential equation:
\[ \frac { dv }{ dx } - 1 = \frac { 1-3v }{ 2v } \] Now, solve for \( \frac { dv }{ dx } \):
\[ \frac { dv }{ dx } = 1 + \frac { 1-3v }{ 2v } \]
\[ \frac { dv }{ dx } = \frac { 2v + (1-3v) }{ 2v } \]
\[ \frac { dv }{ dx } = \frac { 2v+1-3v }{ 2v } \]
\[ \frac { dv }{ dx } = \frac { 1-v }{ 2v } \] Now, separate the variables:
\[ dx = \frac { 2v }{ 1-v } dv \] Integrate both sides:
\[ \int dx = \int \frac { 2v }{ 1-v } dv \] To integrate the right side, we can rewrite the numerator:
\( \frac { 2v }{ 1-v } = \frac { -2(1-v)+2 }{ 1-v } = -2 + \frac { 2 }{ 1-v } \) So, the integral becomes:
\[ \int dx = \int \left( -2 + \frac { 2 }{ 1-v } \right) dv \]
\( x+C_1 = -2v - 2\log|1-v| \) Let \( C = C_1 \) be an arbitrary constant:
\( x+C = -2v - 2\log|1-v| \) Finally, substitute back \( v = x+y \):
\( x+C = -2(x+y) - 2\log|1-(x+y)| \)
\( x+C = -2x-2y - 2\log|1-x-y| \) Rearrange the terms to get the final solution:
\( x + 2x+2y + 2\log|1-x-y| = -C \)
\( 3x+2y + 2\log|1-x-y| = C' \) (where \( C' = -C \)) This is the required solution. This method is effective when the equation can be expressed in terms of a single linear combination of variables.
In simple words: We saw that 'x+y' was common in the equation. So, we replaced it with 'v' to make the equation simpler. After some calculations, we separated the 'x' parts and the 'v' parts, and then integrated them. Finally, we put 'x+y' back for 'v' to get the answer.

🎯 Exam Tip: Always check if a differential equation can be simplified by a substitution like \( v = ax+by \). This often transforms it into a separable variables equation, which is generally easier to solve.

Question 5. \( \frac { dy }{ dx } = \frac { 6x-2y-7 }{ 2x+3y-6 } \)
Answer: The given differential equation is of the form \( \frac { dy }{ dx } = \frac { a_1x+b_1y+c_1 }{ a_2x+b_2y+c_2 } \). We check if \( \frac { a_1 }{ a_2 } = \frac { b_1 }{ b_2 } \). Here \( a_1=6, b_1=-2, a_2=2, b_2=3 \). \( \frac { 6 }{ 2 } = 3 \) and \( \frac { -2 }{ 3 } \). Since \( 3 \ne \frac { -2 }{ 3 } \), this is a non-homogeneous equation where the condition for substitution \( v = ax+by \) is not directly met. We make the substitution \( x = X+h \) and \( y = Y+k \), so \( \frac { dy }{ dx } = \frac { dY }{ dX } \).
\[ \frac { dY }{ dX } = \frac { 6(X+h)-2(Y+k)-7 }{ 2(X+h)+3(Y+k)-6 } \]
\[ \frac { dY }{ dX } = \frac { 6X-2Y+(6h-2k-7) }{ 2X+3Y+(2h+3k-6) } \] For the equation to become homogeneous, the constant terms in the numerator and denominator must be zero:
\( 6h-2k-7 = 0 \) (Equation 1)
\( 2h+3k-6 = 0 \) (Equation 2) Multiply Equation 2 by 3: \( 6h+9k-18=0 \) (Equation 3) Subtract Equation 1 from Equation 3:
\( (6h+9k-18) - (6h-2k-7) = 0 \)
\( 6h+9k-18-6h+2k+7 = 0 \)
\( 11k-11 = 0 \implies 11k=11 \implies k=1 \) Substitute \( k=1 \) into Equation 2:
\( 2h+3(1)-6 = 0 \)
\( 2h+3-6 = 0 \)
\( 2h-3 = 0 \implies 2h=3 \implies h = \frac { 3 }{ 2 } \) So, \( h = \frac { 3 }{ 2 } \) and \( k = 1 \). The differential equation becomes homogeneous:
\[ \frac { dY }{ dX } = \frac { 6X-2Y }{ 2X+3Y } \] Now, we substitute \( Y = vX \), which means \( \frac { dY }{ dX } = v + X \frac { dv }{ dX } \).
\( v + X \frac { dv }{ dX } = \frac { 6X-2vX }{ 2X+3vX } \)
\( v + X \frac { dv }{ dX } = \frac { X(6-2v) }{ X(2+3v) } \)
\( v + X \frac { dv }{ dX } = \frac { 6-2v }{ 2+3v } \)
\( X \frac { dv }{ dX } = \frac { 6-2v }{ 2+3v } - v \)
\( X \frac { dv }{ dX } = \frac { (6-2v) - v(2+3v) }{ 2+3v } \)
\( X \frac { dv }{ dX } = \frac { 6-2v-2v-3v^2 }{ 2+3v } \)
\( X \frac { dv }{ dX } = \frac { 6-4v-3v^2 }{ 2+3v } \) Now, separate the variables:
\[ \frac { dX }{ X } = \frac { (2+3v) }{ (6-4v-3v^2) } dv \] We can factor out -1 from the denominator to make it easier for integration:
\[ \frac { dX }{ X } = - \frac { (2+3v) }{ (3v^2+4v-6) } dv \] Integrate both sides:
\[ \int \frac { dX }{ X } = \int - \frac { (2+3v) }{ (3v^2+4v-6) } dv \] The derivative of the denominator \( 3v^2+4v-6 \) is \( 6v+4 \). We can rewrite the numerator \( (2+3v) \) as \( \frac { 1 }{ 2 } (6v+4) \).
\[ \int \frac { dX }{ X } = - \frac { 1 }{ 2 } \int \frac { (6v+4) }{ (3v^2+4v-6) } dv \]
\( \log|X| = - \frac { 1 }{ 2 } \log|3v^2+4v-6| + \log c \) Multiply by 2:
\( 2 \log|X| = - \log|3v^2+4v-6| + 2 \log c \)
\( \log(X^2) = \log \left( \frac { c^2 }{ 3v^2+4v-6 } \right) \)
\( X^2 = \frac { C }{ 3v^2+4v-6 } \) (where \( C = c^2 \))
\( X^2 (3v^2+4v-6) = C \) Substitute \( v = \frac { Y }{ X } \):
\( X^2 \left( 3\left(\frac{Y}{X}\right)^2 + 4\left(\frac{Y}{X}\right) - 6 \right) = C \)
\( X^2 \left( \frac{3Y^2}{X^2} + \frac{4Y}{X} - 6 \right) = C \)
\( 3Y^2 + 4XY - 6X^2 = C \) Finally, substitute back \( X = x-h = x-\frac { 3 }{ 2 } \) and \( Y = y-k = y-1 \):
\( 3(y-1)^2 + 4\left(x-\frac{3}{2}\right)(y-1) - 6\left(x-\frac{3}{2}\right)^2 = C \) This is the general solution for the given differential equation. This method is crucial for solving non-homogeneous linear differential equations.
In simple words: First, we shifted the coordinates by finding 'h' and 'k' to make the equation simpler (homogeneous). Then we used a substitution with 'v' to separate the 'X' and 'v' parts, and integrated them. Finally, we put the original 'x' and 'y' back to get the final solution.

🎯 Exam Tip: When dealing with non-homogeneous equations, always verify the \( \frac { a_1 }{ a_2 } = \frac { b_1 }{ b_2 } \) condition first. If it's not met, use the \( x=X+h, y=Y+k \) substitution to transform it into a homogeneous equation.