RBSE Solutions Class 12 Maths Chapter 12 Differential Equation Exercise 12.5

Question 1. \( (x + y)² \frac { dy }{ dx} = a² \)
Answer: Let the given differential equation be: \( (x + y)² \frac { dy }{ dx} = a² \) ... (1)
To solve this, we will substitute \( x + y = u \). Now, differentiate both sides with respect to \( x \):
\( 1 + \frac{dy}{dx} = \frac{du}{dx} \)
\( \implies \frac{dy}{dx} = \frac{du}{dx} - 1 \)
Substitute \( x + y = u \) and \( \frac{dy}{dx} = \frac{du}{dx} - 1 \) into equation (1):
\( u² \left( \frac{du}{dx} - 1 \right) = a² \)
\( \implies u² \frac{du}{dx} - u² = a² \)
\( \implies u² \frac{du}{dx} = a² + u² \)
Separate the variables to integrate:
\( \frac{u²}{a² + u²} du = dx \)
Rewrite the left side for easier integration:
\( \frac{u² + a² - a²}{a² + u²} du = dx \)
\( \implies \left( 1 - \frac{a²}{u² + a²} \right) du = dx \)
Integrate both sides:
\( \int \left( 1 - \frac{a²}{u² + a²} \right) du = \int dx \)
\( \implies \int du - a² \int \frac{1}{u² + a²} du = \int dx \)
Recall the standard integral \( \int \frac{1}{x² + a²} dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \). Applying this:
\( u - a² \left( \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) \right) = x + C \)
\( \implies u - a \tan^{-1} \left( \frac{u}{a} \right) = x + C \)
Now, substitute back \( u = x + y \):
\( (x + y) - a \tan^{-1} \left( \frac{x + y}{a} \right) = x + C \)
Simplify by canceling \( x \) from both sides:
\( y - a \tan^{-1} \left( \frac{x + y}{a} \right) = C \)
Rearrange to solve for \( y \):
\( y = C + a \tan^{-1} \left( \frac{x + y}{a} \right) \)
This can also be written as:
\( \frac{y - C}{a} = \tan^{-1} \left( \frac{x + y}{a} \right) \)
\( \implies \tan \left( \frac{y - C}{a} \right) = \frac{x + y}{a} \)
\( \implies a \tan \left( \frac{y - C}{a} \right) = x + y \)
So, the final solution is \( y = a \tan \left( \frac{y - C}{a} \right) - x \). A small detail to note is that constant C can absorb the negative sign, so the form \( y = a \tan^{-1} \left( \frac{x + y}{a} \right) + C \) is also valid. In simple words: We changed \( x+y \) to \( u \) to make the equation simpler. Then we separated parts with \( u \) and \( x \) and solved them by integration. Finally, we put \( x+y \) back in place of \( u \) to get the answer.

🎯 Exam Tip: When a differential equation has terms like \( (x+y) \) or \( (x-y) \) appearing multiple times, a substitution like \( u = x+y \) often simplifies it significantly, transforming it into a separable variables form.

Question 2. \( \frac { dy }{ dx} = \frac { 1 }{ x+y+1} \)
Answer: Let the given differential equation be:
\( \frac { dy }{ dx} = \frac { 1 }{ x+y+1} \) ... (1)
To solve this, we substitute \( t = x + y + 1 \).
Differentiate both sides with respect to \( x \):
\( \frac{dt}{dx} = 1 + \frac{dy}{dx} + 0 \)
\( \implies \frac{dy}{dx} = \frac{dt}{dx} - 1 \)
Substitute \( t \) and \( \frac{dy}{dx} \) into equation (1):
\( \frac{dt}{dx} - 1 = \frac{1}{t} \)
\( \implies \frac{dt}{dx} = 1 + \frac{1}{t} \)
\( \implies \frac{dt}{dx} = \frac{t + 1}{t} \)
Separate the variables to integrate:
\( \frac{t}{t+1} dt = dx \)
Rewrite the left side for easier integration:
\( \frac{t+1-1}{t+1} dt = dx \)
\( \implies \left( 1 - \frac{1}{t+1} \right) dt = dx \)
Integrate both sides:
\( \int \left( 1 - \frac{1}{t+1} \right) dt = \int dx \)
\( \implies \int dt - \int \frac{1}{t+1} dt = \int dx \)
\( \implies t - \log |t+1| = x + C_1 \)
Now, substitute back \( t = x + y + 1 \):
\( (x + y + 1) - \log |(x + y + 1) + 1| = x + C_1 \)
\( \implies x + y + 1 - \log |x + y + 2| = x + C_1 \)
Simplify by canceling \( x \) from both sides:
\( y + 1 - \log |x + y + 2| = C_1 \)
Rearrange the terms to express the solution:
\( \log |x + y + 2| = y + 1 - C_1 \)
Let \( C = 1 - C_1 \). This combines constants.
\( \log |x + y + 2| = y + C \)
Using the definition of logarithm, \( \log_b a = c \implies b^c = a \):
\( x + y + 2 = e^{y+C} \)
\( \implies x + y + 2 = e^y \cdot e^C \)
Let \( e^C = A \) (where A is a new arbitrary constant).
\( x + y + 2 = A e^y \)
This is the general solution. The constant can be represented in various forms. In simple words: We used a substitution to turn the complex equation into a simpler one. After integrating, we replaced the substituted variable to get the final answer. This method makes it easy to solve equations where \( x \) and \( y \) appear together in a specific way.

🎯 Exam Tip: When you see \( (ax+by+c) \) as a grouping in a differential equation, try substituting \( t = ax+by+c \). This often simplifies the equation, allowing for separation of variables.

Question 3. \( \cos (x + y) dy = dx \)
Answer: Let the given differential equation be:
\( \cos (x + y) dy = dx \)
Rearrange to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{1}{\cos(x+y)} \)
Let \( v = x + y \).
Differentiate \( v \) with respect to \( x \):
\( \frac{dv}{dx} = 1 + \frac{dy}{dx} \)
\( \implies \frac{dy}{dx} = \frac{dv}{dx} - 1 \)
Substitute this into the rearranged equation:
\( \frac{dv}{dx} - 1 = \frac{1}{\cos v} \)
\( \implies \frac{dv}{dx} = 1 + \frac{1}{\cos v} \)
\( \implies \frac{dv}{dx} = \frac{\cos v + 1}{\cos v} \)
Separate the variables to integrate:
\( \frac{\cos v}{1 + \cos v} dv = dx \)
To integrate the left side, we can rewrite the numerator:
\( \frac{1 + \cos v - 1}{1 + \cos v} dv = dx \)
\( \implies \left( 1 - \frac{1}{1 + \cos v} \right) dv = dx \)
We know that \( 1 + \cos v = 2 \cos^2 \left( \frac{v}{2} \right) \). So,
\( 1 - \frac{1}{2 \cos^2 \left( \frac{v}{2} \right)} dv = dx \)
\( \implies \left( 1 - \frac{1}{2} \sec^2 \left( \frac{v}{2} \right) \right) dv = dx \)
Integrate both sides:
\( \int \left( 1 - \frac{1}{2} \sec^2 \left( \frac{v}{2} \right) \right) dv = \int dx \)
\( \implies \int dv - \frac{1}{2} \int \sec^2 \left( \frac{v}{2} \right) dv = \int dx \)
Recall that \( \int \sec^2(ax) dx = \frac{1}{a} \tan(ax) \). Here, \( a = \frac{1}{2} \).
\( \implies v - \frac{1}{2} \left( \frac{1}{1/2} \tan \left( \frac{v}{2} \right) \right) = x + C \)
\( \implies v - \tan \left( \frac{v}{2} \right) = x + C \)
Now substitute back \( v = x + y \):
\( (x + y) - \tan \left( \frac{x + y}{2} \right) = x + C \)
Simplify by canceling \( x \) from both sides:
\( y - \tan \left( \frac{x + y}{2} \right) = C \)
\( \implies y = C + \tan \left( \frac{x + y}{2} \right) \)
This is the general solution for the differential equation. The constant C can be determined if initial conditions are given. In simple words: First, we changed \( x+y \) to \( v \) to simplify the equation. Then we separated the parts with \( v \) and \( x \) so we could integrate them. After integrating, we put \( x+y \) back to get the final answer.

🎯 Exam Tip: Remember to use trigonometric identities like \( 1 + \cos v = 2 \cos^2 \left( \frac{v}{2} \right) \) when integrating expressions involving \( 1 \pm \cos v \) or \( 1 \pm \sin v \). This helps convert them into forms that are easier to integrate.

Question 4. \( e^{x+y} = 1 + \frac { dy }{ dx } \)
Answer: Let the given differential equation be:
\( e^{x+y} = 1 + \frac { dy }{ dx } \) ... (1)
To solve this, we substitute \( v = x + y \).
Differentiate \( v \) with respect to \( x \):
\( \frac{dv}{dx} = 1 + \frac{dy}{dx} \)
Substitute \( v \) and \( \frac{dv}{dx} \) into equation (1):
\( e^v = \frac{dv}{dx} \)
Separate the variables to integrate:
\( e^{-v} dv = dx \)
Integrate both sides:
\( \int e^{-v} dv = \int dx \)
\( \implies -e^{-v} = x + C_1 \)
Now, substitute back \( v = x + y \):
\( -e^{-(x+y)} = x + C_1 \)
We can rearrange the terms to make it clearer:
\( x + e^{-(x+y)} = -C_1 \)
Let \( C = -C_1 \) (where C is a new arbitrary constant).
So, the general solution is \( x + e^{-(x+y)} = C \). This compact form represents the family of solutions. In simple words: We made the equation simpler by replacing \( x+y \) with \( v \). Then, we separated the parts with \( v \) and \( x \) and solved them using integration. Finally, we put \( x+y \) back to get the solution.

🎯 Exam Tip: When you see \( e^{x+y} \) or \( e^{x-y} \) in a differential equation, a substitution like \( v = x+y \) or \( v = x-y \) often simplifies the equation to a separable form. This helps in integrating the equation easily.

Question 5. \( (x + y) (dx - dy) = dx + dy \)
Answer: Let the given differential equation be:
\( (x + y) (dx - dy) = dx + dy \)
This equation can be rewritten by dividing both sides by \( dx \):
\( (x + y) \left( 1 - \frac{dy}{dx} \right) = 1 + \frac{dy}{dx} \)
To solve this, we substitute \( v = x + y \).
Differentiate \( v \) with respect to \( x \):
\( \frac{dv}{dx} = 1 + \frac{dy}{dx} \)
From this, we can also write \( \frac{dy}{dx} = \frac{dv}{dx} - 1 \).
Now, let's substitute these into the equation:
\( v \left( 1 - \left( \frac{dv}{dx} - 1 \right) \right) = \frac{dv}{dx} \)
\( \implies v \left( 1 - \frac{dv}{dx} + 1 \right) = \frac{dv}{dx} \)
\( \implies v \left( 2 - \frac{dv}{dx} \right) = \frac{dv}{dx} \)
\( \implies 2v - v \frac{dv}{dx} = \frac{dv}{dx} \)
Move all terms with \( \frac{dv}{dx} \) to one side:
\( 2v = \frac{dv}{dx} + v \frac{dv}{dx} \)
\( \implies 2v = \frac{dv}{dx} (1 + v) \)
Separate the variables to integrate:
\( \frac{1+v}{2v} dv = dx \)
\( \implies \left( \frac{1}{2v} + \frac{v}{2v} \right) dv = dx \)
\( \implies \left( \frac{1}{2v} + \frac{1}{2} \right) dv = dx \)
Integrate both sides:
\( \int \left( \frac{1}{2v} + \frac{1}{2} \right) dv = \int dx \)
\( \implies \frac{1}{2} \int \frac{1}{v} dv + \frac{1}{2} \int dv = \int dx \)
\( \implies \frac{1}{2} \log |v| + \frac{1}{2} v = x + C_1 \)
Multiply by 2 to clear the fractions:
\( \log |v| + v = 2x + 2C_1 \)
Let \( C = 2C_1 \) (a new arbitrary constant).
\( \log |v| + v = 2x + C \)
Now, substitute back \( v = x + y \):
\( \log |x + y| + (x + y) = 2x + C \)
Rearrange the terms:
\( \log |x + y| = 2x - x - y + C \)
\( \implies \log |x + y| = x - y + C \)
This is the general solution for the differential equation. The natural logarithm makes it challenging to explicitly solve for y. In simple words: We used a substitution for \( x+y \) and then separated the equation so we could integrate both sides. This helped us solve the equation and find the relationship between \( x \) and \( y \).

🎯 Exam Tip: For equations like this, where \( dx \) and \( dy \) appear together in complex ways, dividing by \( dx \) to get \( \frac{dy}{dx} \) is the first critical step. Then, look for common groups like \( x+y \) for substitution.

Question 6. \( \frac { dy }{ dx} = \frac { x+y+1 }{ x+y} \)
Answer: Let the given differential equation be:
\( \frac { dy }{ dx} = \frac { x+y+1 }{ x+y} \) ... (1)
To solve this, we substitute \( v = x + y \).
Differentiate \( v \) with respect to \( x \):
\( \frac{dv}{dx} = 1 + \frac{dy}{dx} \)
\( \implies \frac{dy}{dx} = \frac{dv}{dx} - 1 \)
Substitute \( v \) and \( \frac{dy}{dx} \) into equation (1):
\( \frac{dv}{dx} - 1 = \frac{v + 1}{v} \)
\( \implies \frac{dv}{dx} = 1 + \frac{v + 1}{v} \)
\( \implies \frac{dv}{dx} = \frac{v + (v + 1)}{v} \)
\( \implies \frac{dv}{dx} = \frac{2v + 1}{v} \)
Separate the variables to integrate:
\( \frac{v}{2v+1} dv = dx \)
To integrate the left side, we can rewrite the numerator:
\( \frac{1}{2} \frac{2v}{2v+1} dv = dx \)
\( \implies \frac{1}{2} \frac{2v+1-1}{2v+1} dv = dx \)
\( \implies \frac{1}{2} \left( 1 - \frac{1}{2v+1} \right) dv = dx \)
Integrate both sides:
\( \int \frac{1}{2} \left( 1 - \frac{1}{2v+1} \right) dv = \int dx \)
\( \implies \frac{1}{2} \left( \int 1 dv - \int \frac{1}{2v+1} dv \right) = \int dx \)
\( \implies \frac{1}{2} \left( v - \frac{\log|2v+1|}{2} \right) = x + C_1 \)
Multiply by 2 to clear the fraction:
\( v - \frac{1}{2} \log|2v+1| = 2x + 2C_1 \)
Now, substitute back \( v = x + y \):
\( (x + y) - \frac{1}{2} \log|2(x + y)+1| = 2x + 2C_1 \)
Rearrange the terms:
\( y - \frac{1}{2} \log|2x + 2y + 1| = 2x - x + 2C_1 \)
\( \implies y - \frac{1}{2} \log|2x + 2y + 1| = x + 2C_1 \)
Multiply by 2:
\( 2y - \log|2x + 2y + 1| = 2x + 4C_1 \)
Let \( C = 4C_1 \) (a new arbitrary constant).
\( 2y - \log|2x + 2y + 1| = 2x + C \)
This is the general solution. It shows how the variables relate to each other after integrating. In simple words: We used a substitution for \( x+y \) to make the equation simpler. Then we separated the parts with the new variable and \( x \) to integrate them. Finally, we replaced \( x+y \) back to find the complete solution.

🎯 Exam Tip: Always pay attention to the coefficients when integrating terms like \( \int \frac{1}{ax+b} dx = \frac{1}{a} \log|ax+b| \). For \( \int \frac{1}{2v+1} dv \), the \( \frac{1}{2} \) factor is crucial.

Question 7. \( x + y = \sin^{-1} \left( \frac {dy }{dx} \right) \)
Answer: Let the given differential equation be:
\( x + y = \sin^{-1} \left( \frac {dy }{dx} \right) \)
This can be rewritten as:
\( \sin(x + y) = \frac{dy}{dx} \) ... (1)
To solve this, we substitute \( u = x + y \).
Differentiate \( u \) with respect to \( x \):
\( \frac{du}{dx} = 1 + \frac{dy}{dx} \)
\( \implies \frac{dy}{dx} = \frac{du}{dx} - 1 \)
Substitute \( u \) and \( \frac{dy}{dx} \) into equation (1):
\( \sin u = \frac{du}{dx} - 1 \)
\( \implies \frac{du}{dx} = 1 + \sin u \)
Separate the variables to integrate:
\( \frac{1}{1 + \sin u} du = dx \)
To integrate the left side, we multiply the numerator and denominator by \( (1 - \sin u) \):
\( \frac{1 - \sin u}{(1 + \sin u)(1 - \sin u)} du = dx \)
\( \implies \frac{1 - \sin u}{1 - \sin^2 u} du = dx \)
Using the identity \( 1 - \sin^2 u = \cos^2 u \):
\( \frac{1 - \sin u}{\cos^2 u} du = dx \)
\( \implies \left( \frac{1}{\cos^2 u} - \frac{\sin u}{\cos^2 u} \right) du = dx \)
\( \implies \left( \sec^2 u - \frac{\sin u}{\cos u} \cdot \frac{1}{\cos u} \right) du = dx \)
\( \implies (\sec^2 u - \tan u \sec u) du = dx \)
Integrate both sides:
\( \int (\sec^2 u - \tan u \sec u) du = \int dx \)
\( \implies \int \sec^2 u du - \int \tan u \sec u du = \int dx \)
Recall the standard integrals: \( \int \sec^2 u du = \tan u \) and \( \int \tan u \sec u du = \sec u \).
\( \implies \tan u - \sec u = x + C_1 \)
Now, substitute back \( u = x + y \):
\( \tan (x + y) - \sec (x + y) = x + C_1 \)
This is the general solution to the differential equation. The constant \( C_1 \) can be replaced with \( C \) for simplicity. In simple words: We changed the given equation into a simpler form by replacing \( x+y \) with \( u \). Then we separated the parts with \( u \) and \( x \) and integrated them. Finally, we put \( x+y \) back to find the solution.

🎯 Exam Tip: When \( \sin^{-1} \) is present, immediately convert it to the direct sine function. Then, use trigonometric identities like \( \sin^2 u + \cos^2 u = 1 \) to simplify the integrand before integration.

Question 8. \( \frac { dy }{ dx} = \frac { 1 }{ x-y } + 1 \)
Answer: Let the given differential equation be:
\( \frac { dy }{ dx} = \frac { 1 }{ x-y } + 1 \) ... (1)
To solve this, we substitute \( v = x - y \).
Differentiate \( v \) with respect to \( x \):
\( \frac{dv}{dx} = 1 - \frac{dy}{dx} \)
\( \implies \frac{dy}{dx} = 1 - \frac{dv}{dx} \)
Substitute \( v \) and \( \frac{dy}{dx} \) into equation (1):
\( 1 - \frac{dv}{dx} = \frac{1}{v} + 1 \)
Cancel 1 from both sides:
\( - \frac{dv}{dx} = \frac{1}{v} \)
Separate the variables to integrate:
\( -v dv = dx \)
Integrate both sides:
\( \int -v dv = \int dx \)
\( \implies -\frac{v^2}{2} = x + C_1 \)
Multiply by 2:
\( -v^2 = 2x + 2C_1 \)
Now, substitute back \( v = x - y \):
\( -(x - y)² = 2x + 2C_1 \)
Rearrange the terms:
\( 2x + (x - y)² = -2C_1 \)
Let \( C = -2C_1 \) (a new arbitrary constant).
So, the general solution is \( 2x + (x - y)² = C \). This represents the family of curves that satisfy the differential equation. In simple words: We simplified the equation by replacing \( x-y \) with \( v \). Then we separated the parts with \( v \) and \( x \) and integrated them. Finally, we put \( x-y \) back to get the complete solution.

🎯 Exam Tip: When using substitution for \( x \pm y \), be careful with the signs when differentiating. \( \frac{d}{dx}(x-y) = 1 - \frac{dy}{dx} \) is different from \( \frac{d}{dx}(x+y) = 1 + \frac{dy}{dx} \).

Question 9. \( \frac {dy }{dx} = \sec (x + y) \)
Answer: Let the given differential equation be:
\( \frac {dy }{dx} = \sec (x + y) \) ... (1)
To solve this, we substitute \( v = x + y \).
Differentiate \( v \) with respect to \( x \):
\( \frac{dv}{dx} = 1 + \frac{dy}{dx} \)
\( \implies \frac{dy}{dx} = \frac{dv}{dx} - 1 \)
Substitute \( v \) and \( \frac{dy}{dx} \) into equation (1):
\( \frac{dv}{dx} - 1 = \sec v \)
\( \implies \frac{dv}{dx} = 1 + \sec v \)
Separate the variables to integrate:
\( \frac{1}{1 + \sec v} dv = dx \)
Rewrite \( \sec v \) as \( \frac{1}{\cos v} \):
\( \frac{1}{1 + \frac{1}{\cos v}} dv = dx \)
\( \implies \frac{1}{\frac{\cos v + 1}{\cos v}} dv = dx \)
\( \implies \frac{\cos v}{\cos v + 1} dv = dx \)
To integrate the left side, we rewrite the numerator:
\( \frac{\cos v + 1 - 1}{\cos v + 1} dv = dx \)
\( \implies \left( 1 - \frac{1}{\cos v + 1} \right) dv = dx \)
We use the identity \( \cos v + 1 = 2 \cos^2 \left( \frac{v}{2} \right) \):
\( \implies \left( 1 - \frac{1}{2 \cos^2 \left( \frac{v}{2} \right)} \right) dv = dx \)
\( \implies \left( 1 - \frac{1}{2} \sec^2 \left( \frac{v}{2} \right) \right) dv = dx \)
Integrate both sides:
\( \int \left( 1 - \frac{1}{2} \sec^2 \left( \frac{v}{2} \right) \right) dv = \int dx \)
\( \implies v - \frac{1}{2} \left( \frac{1}{1/2} \tan \left( \frac{v}{2} \right) \right) = x + C \)
\( \implies v - \tan \left( \frac{v}{2} \right) = x + C \)
Now, substitute back \( v = x + y \):
\( (x + y) - \tan \left( \frac{x + y}{2} \right) = x + C \)
Simplify by canceling \( x \) from both sides:
\( y - \tan \left( \frac{x + y}{2} \right) = C \)
\( \implies y = C + \tan \left( \frac{x + y}{2} \right) \)
This is the general solution for the differential equation. The constant C reflects the family of solutions. In simple words: We used a substitution for \( x+y \) to make the equation simpler. Then we separated the parts with the new variable and \( x \) and integrated them using trigonometric identities. Finally, we replaced \( x+y \) back to find the complete solution.

🎯 Exam Tip: Remember that \( \sec v = \frac{1}{\cos v} \). Converting secant to cosine often helps in simplifying expressions for integration, especially when dealing with sums or differences like \( 1 + \sec v \).

Question 10. \( \frac {dy}{dx} = \frac{(x-y)+3}{2(x-y)+5} \)
Answer: Let the given differential equation be:
\( \frac {dy}{dx} = \frac{(x-y)+3}{2(x-y)+5} \) ... (1)
To solve this, we substitute \( v = x - y \).
Differentiate \( v \) with respect to \( x \):
\( \frac{dv}{dx} = 1 - \frac{dy}{dx} \)
\( \implies \frac{dy}{dx} = 1 - \frac{dv}{dx} \)
Substitute \( v \) and \( \frac{dy}{dx} \) into equation (1):
\( 1 - \frac{dv}{dx} = \frac{v+3}{2v+5} \)
\( \implies \frac{dv}{dx} = 1 - \frac{v+3}{2v+5} \)
Combine the terms on the right side:
\( \frac{dv}{dx} = \frac{(2v+5) - (v+3)}{2v+5} \)
\( \implies \frac{dv}{dx} = \frac{2v+5-v-3}{2v+5} \)
\( \implies \frac{dv}{dx} = \frac{v+2}{2v+5} \)
Separate the variables to integrate:
\( \frac{2v+5}{v+2} dv = dx \)
To integrate the left side, we rewrite the numerator:
\( \frac{2(v+2)+1}{v+2} dv = dx \)
\( \implies \left( \frac{2(v+2)}{v+2} + \frac{1}{v+2} \right) dv = dx \)
\( \implies \left( 2 + \frac{1}{v+2} \right) dv = dx \)
Integrate both sides:
\( \int \left( 2 + \frac{1}{v+2} \right) dv = \int dx \)
\( \implies 2 \int dv + \int \frac{1}{v+2} dv = \int dx \)
\( \implies 2v + \log|v+2| = x + C \)
Now, substitute back \( v = x - y \):
\( 2(x - y) + \log|(x - y)+2| = x + C \)
Simplify and rearrange the terms:
\( 2x - 2y + \log|x - y + 2| = x + C \)
\( \implies \log|x - y + 2| = x + 2y - 2x + C \)
\( \implies \log|x - y + 2| = 2y - x + C \)
This is the general solution for the differential equation. The constant C represents the family of solutions. In simple words: We used a substitution for \( x-y \) to simplify the equation. Then we separated the parts with the new variable and \( x \) and integrated them. Finally, we put \( x-y \) back to get the complete solution for the equation.

🎯 Exam Tip: When integrating rational functions like \( \frac{2v+5}{v+2} \), always try to perform polynomial division or rewrite the numerator in terms of the denominator to simplify it into an easily integrable form.