Samacheer Kalvi Class 12 Maths Solutions Chapter 10 Ordinary Differential Equations Exercise 10.6

Question 1. Solve the following differential equations. \( [x + y \cos(\frac { y }{ x })] dx = x \cos (\frac { y }{ x }) dy \)
Answer: The given differential equation is \( [x + y \cos(\frac { y }{ x })] dx = x \cos (\frac { y }{ x }) dy \).
We can rearrange this equation to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{x + y \cos(\frac{y}{x})}{x \cos(\frac{y}{x})} \)
\( \frac{dy}{dx} = \frac{x}{x \cos(\frac{y}{x})} + \frac{y \cos(\frac{y}{x})}{x \cos(\frac{y}{x})} \)
\( \frac{dy}{dx} = \frac{1}{\cos(\frac{y}{x})} + \frac{y}{x} \)
\( \frac{dy}{dx} = \sec(\frac{y}{x}) + \frac{y}{x} \)... (1)
This is a homogeneous differential equation because the terms involve \( \frac{y}{x} \).
Let's substitute \( y = vx \).
Then, \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into equation (1):
\( v + x \frac{dv}{dx} = \sec(v) + v \)
\( x \frac{dv}{dx} = \sec(v) \)
Now, we separate the variables:
\( \frac{dv}{\sec(v)} = \frac{dx}{x} \)
\( \cos(v) dv = \frac{dx}{x} \)
Next, integrate both sides:
\( \int \cos(v) dv = \int \frac{dx}{x} \)
\( \sin(v) = \log|x| + \log|c| \) (where `c` is the integration constant)
We can combine the logarithmic terms:
\( \sin(v) = \log|cx| \)
Finally, substitute back \( v = \frac{y}{x} \):
\( \sin(\frac{y}{x}) = \log|cx| \)
This is the required general solution for the differential equation. The constant `c` combines the initial conditions of the problem.
In simple words: First, we change the given equation into a form where we can see if it's "homogeneous," which means it has a special pattern. Then, we use a trick by replacing `y` with `vx` to make it easier to solve. After that, we separate the parts with `v` and `x` and integrate them to get the final answer.

🎯 Exam Tip: Always check if a differential equation is homogeneous by rewriting \( \frac{dy}{dx} \) in terms of \( \frac{y}{x} \) only. If it is, the substitution \( y = vx \) is a standard and effective method.

Question 2. Solve \( (x³ + y³) dy = x²y dx \)
Answer: The given differential equation is \( (x³ + y³) dy = x²y dx \).
First, we express \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{x²y}{x³ + y³} \)
This is a homogeneous differential equation since all terms have the same total degree (3).
Let's substitute \( y = vx \).
Then, \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into the equation:
\( v + x \frac{dv}{dx} = \frac{x²(vx)}{x³ + (vx)³} \)
\( v + x \frac{dv}{dx} = \frac{vx³}{x³ + v³x³} \)
\( v + x \frac{dv}{dx} = \frac{vx³}{x³(1 + v³)} \)
\( v + x \frac{dv}{dx} = \frac{v}{1 + v³} \)
Now, we isolate \( x \frac{dv}{dx} \):
\( x \frac{dv}{dx} = \frac{v}{1 + v³} - v \)
\( x \frac{dv}{dx} = \frac{v - v(1 + v³)}{1 + v³} \)
\( x \frac{dv}{dx} = \frac{v - v - v⁴}{1 + v³} \)
\( x \frac{dv}{dx} = \frac{-v⁴}{1 + v³} \)
Separate the variables:
\( \frac{1 + v³}{-v⁴} dv = \frac{dx}{x} \)
\( - \frac{1 + v³}{v⁴} dv = \frac{dx}{x} \)
\( - (\frac{1}{v⁴} + \frac{v³}{v⁴}) dv = \frac{dx}{x} \)
\( - (v^{-4} + v^{-1}) dv = \frac{dx}{x} \)
Integrate both sides:
\( - \int (v^{-4} + v^{-1}) dv = \int \frac{dx}{x} \)
\( - (\frac{v^{-3}}{-3} + \log|v|) = \log|x| + C \)
\( \frac{1}{3v³} - \log|v| = \log|x| + C \)
Rearrange terms:
\( \frac{1}{3v³} = \log|x| + \log|v| + C \)
\( \frac{1}{3v³} = \log|xv| + C \)
Substitute back \( v = \frac{y}{x} \):
\( \frac{1}{3(\frac{y}{x})³} = \log|x(\frac{y}{x})| + C \)
\( \frac{x³}{3y³} = \log|y| + C \)
This is the implicit general solution. The constant `C` is crucial for the solution.
In simple words: We check if the equation follows a "homogeneous" pattern, meaning all its parts are of the same degree. Then we use a substitution to simplify it, separate the variables, and integrate each part. Finally, we put the original variables back into the answer.

🎯 Exam Tip: Remember to simplify the terms after substitution, especially when powers of `x` cancel out, before separating variables for integration.

Question 3. Solve \( ye^{\frac{x}{y}} dx = (xe^{\frac{x}{y}} + y)dy \)
Answer: The given differential equation is \( ye^{\frac{x}{y}} dx = (xe^{\frac{x}{y}} + y)dy \).
This time, it's easier to express \( \frac{dx}{dy} \) because the exponential term has \( \frac{x}{y} \):
\( \frac{dx}{dy} = \frac{xe^{\frac{x}{y}} + y}{ye^{\frac{x}{y}}} \)
\( \frac{dx}{dy} = \frac{xe^{\frac{x}{y}}}{ye^{\frac{x}{y}}} + \frac{y}{ye^{\frac{x}{y}}} \)
\( \frac{dx}{dy} = \frac{x}{y} + e^{-\frac{x}{y}} \)
This is a homogeneous differential equation.
Let's substitute \( x = vy \).
Then, \( \frac{dx}{dy} = v + y \frac{dv}{dy} \).
Substitute these into the equation:
\( v + y \frac{dv}{dy} = v + e^{-v} \)
\( y \frac{dv}{dy} = e^{-v} \)
Now, we separate the variables:
\( \frac{dv}{e^{-v}} = \frac{dy}{y} \)
\( e^v dv = \frac{dy}{y} \)
Next, integrate both sides:
\( \int e^v dv = \int \frac{dy}{y} \)
\( e^v = \log|y| + C \)
Finally, substitute back \( v = \frac{x}{y} \):
\( e^{\frac{x}{y}} = \log|y| + C \)
This is the general solution for the given differential equation. The constant `C` can be determined with initial conditions.
In simple words: We change the equation to a form where `x` is related to `y` and `dx/dy` is used. Then, we recognize it as a "homogeneous" equation and use a special trick by replacing `x` with `vy`. After separating the parts, we integrate them to find the solution.

🎯 Exam Tip: When a homogeneous equation involves \( \frac{x}{y} \) terms, substituting \( x=vy \) and finding \( \frac{dx}{dy} \) is often more efficient than the usual \( y=vx \) approach.

Question 4. Solve \( 2xy dx + (x² + 2y²)dy = 0 \)
Answer: The given differential equation is \( 2xy dx + (x² + 2y²)dy = 0 \).
First, let's write it in the form \( \frac{dy}{dx} \):
\( (x² + 2y²)dy = -2xy dx \)
\( \frac{dy}{dx} = \frac{-2xy}{x² + 2y²} \)
This is a homogeneous differential equation.
Let's substitute \( y = vx \).
Then, \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into the equation:
\( v + x \frac{dv}{dx} = \frac{-2x(vx)}{x² + 2(vx)²} \)
\( v + x \frac{dv}{dx} = \frac{-2vx²}{x² + 2v²x²} \)
\( v + x \frac{dv}{dx} = \frac{-2vx²}{x²(1 + 2v²)} \)
\( v + x \frac{dv}{dx} = \frac{-2v}{1 + 2v²} \)
Now, isolate \( x \frac{dv}{dx} \):
\( x \frac{dv}{dx} = \frac{-2v}{1 + 2v²} - v \)
\( x \frac{dv}{dx} = \frac{-2v - v(1 + 2v²)}{1 + 2v²} \)
\( x \frac{dv}{dx} = \frac{-2v - v - 2v³}{1 + 2v²} \)
\( x \frac{dv}{dx} = \frac{-3v - 2v³}{1 + 2v²} \)
\( x \frac{dv}{dx} = \frac{-(3v + 2v³)}{1 + 2v²} \)
Separate the variables:
\( \frac{1 + 2v²}{3v + 2v³} dv = - \frac{dx}{x} \)
We can rewrite the denominator as \( v(3 + 2v²) \).
Consider the left side integral: \( \int \frac{1 + 2v²}{v(3 + 2v³)} dv \).
Let \( u = 3v + 2v³ \). Then \( du = (3 + 6v²) dv \). This does not match directly.
Let's try multiplying the numerator by 3:
\( \frac{1}{3} \int \frac{3(1 + 2v²)}{v(3 + 2v³)} dv = - \int \frac{dx}{x} \)
\( \frac{1}{3} \int \frac{3 + 6v²}{3v + 2v³} dv = - \int \frac{dx}{x} \)
Now, the numerator \( (3 + 6v²) \) is the derivative of the denominator \( (3v + 2v³) \).
\( \frac{1}{3} \log|3v + 2v³| = - \log|x| + \log|C| \)
\( \log|3v + 2v³|^{1/3} = \log|\frac{C}{x}| \)
\( (3v + 2v³)^{1/3} = \frac{C}{x} \)
Cube both sides:
\( 3v + 2v³ = \frac{C³}{x³} \)
Let \( K = C³ \) be a new constant.
\( 3v + 2v³ = \frac{K}{x³} \)
Substitute back \( v = \frac{y}{x} \):
\( 3(\frac{y}{x}) + 2(\frac{y}{x})³ = \frac{K}{x³} \)
\( \frac{3y}{x} + \frac{2y³}{x³} = \frac{K}{x³} \)
Multiply by \( x³ \) to clear denominators:
\( 3yx² + 2y³ = K \)
This is the general solution. This shows that careful algebraic manipulation can simplify integration.
In simple words: We first rewrite the equation to find `dy/dx`. Since it's a special type called "homogeneous," we replace `y` with `vx`. Then, we separate all the `v` terms from the `x` terms. After integrating both sides, we put `y/x` back in place of `v` to get the final solution.

🎯 Exam Tip: When integrating \( \frac{f'(x)}{f(x)} \), the result is \( \log|f(x)| \). Look for opportunities to use this pattern by adjusting constants in the numerator.

Question 5. Solve \( (y² – 2xy) dx = (x² – 2xy) dy \)
Answer: The given differential equation is \( (y² – 2xy) dx = (x² – 2xy) dy \).
First, we express \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{y² – 2xy}{x² – 2xy} \)
This is a homogeneous differential equation because all terms have a total degree of 2.
Let's substitute \( y = vx \).
Then, \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into the equation:
\( v + x \frac{dv}{dx} = \frac{(vx)² – 2x(vx)}{x² – 2x(vx)} \)
\( v + x \frac{dv}{dx} = \frac{v²x² – 2vx²}{x² – 2vx²} \)
\( v + x \frac{dv}{dx} = \frac{x²(v² – 2v)}{x²(1 – 2v)} \)
\( v + x \frac{dv}{dx} = \frac{v² – 2v}{1 – 2v} \)
Now, isolate \( x \frac{dv}{dx} \):
\( x \frac{dv}{dx} = \frac{v² – 2v}{1 – 2v} - v \)
\( x \frac{dv}{dx} = \frac{v² – 2v - v(1 – 2v)}{1 – 2v} \)
\( x \frac{dv}{dx} = \frac{v² – 2v - v + 2v²}{1 – 2v} \)
\( x \frac{dv}{dx} = \frac{3v² – 3v}{1 – 2v} \)
Separate the variables:
\( \frac{1 – 2v}{3v² – 3v} dv = \frac{dx}{x} \)
We can factor the denominator as \( 3v(v-1) \) or \( 3(v^2 - v) \).
\( \int \frac{1 – 2v}{3(v² – v)} dv = \int \frac{dx}{x} \)
Notice that the derivative of \( v² – v \) is \( 2v – 1 = -(1 – 2v) \).
So, we can write:
\( - \frac{1}{3} \int \frac{2v – 1}{v² – v} dv = \int \frac{dx}{x} \)
This matches the form \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| \).
\( - \frac{1}{3} \log|v² – v| = \log|x| + \log|C| \)
\( \log|v² – v|^{-1/3} = \log|Cx| \)
\( (v² – v)^{-1/3} = Cx \)
\( \frac{1}{(v² – v)^{1/3}} = Cx \)
\( 1 = Cx (v² – v)^{1/3} \)
Cube both sides:
\( 1 = C³x³ (v² – v) \)
Let \( K = C³ \) be a new constant.
\( 1 = Kx³(v² – v) \)
Substitute back \( v = \frac{y}{x} \):
\( 1 = Kx³((\frac{y}{x})² – \frac{y}{x}) \)
\( 1 = Kx³(\frac{y²}{x²} – \frac{y}{x}) \)
\( 1 = Kx³(\frac{y² – yx}{x²}) \)
\( 1 = Kx(y² – yx) \)
\( 1 = Kxy(y – x) \)
Alternatively, \( Kxy(y - x) = 1 \) is the solution. This process requires careful factorisation and logarithmic integration.
In simple words: First, we rearrange the equation to find `dy/dx` and notice that all terms have the same power sum, making it "homogeneous." We then substitute `y` with `vx` to simplify it. After separating the variables, we integrate both sides using a special rule for fractions where the top is almost the derivative of the bottom. Finally, we put `y/x` back in place of `v` to get the answer.

🎯 Exam Tip: Pay close attention to the signs and constant factors when using the \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| \) integration method. A common mistake is missing a negative sign or a numerical factor.

Question 6. Solve \( x \frac { dy }{ dx } = y – x \cos²(\frac { y }{ x }) \)
Answer: The given differential equation is \( x \frac { dy }{ dx } = y – x \cos²(\frac { y }{ x }) \).
First, we express \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{y}{x} – \cos²(\frac { y }{ x }) \)
This is a homogeneous differential equation.
Let's substitute \( y = vx \).
Then, \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into the equation:
\( v + x \frac{dv}{dx} = v – \cos²(v) \)
\( x \frac{dv}{dx} = – \cos²(v) \)
Now, we separate the variables:
\( \frac{dv}{-\cos²(v)} = \frac{dx}{x} \)
\( -\sec²(v) dv = \frac{dx}{x} \)
Next, integrate both sides:
\( \int -\sec²(v) dv = \int \frac{dx}{x} \)
\( -\tan(v) = \log|x| + C \)
We can write the constant \( C \) as \( -\log|C_1| \) for convenience.
\( -\tan(v) = \log|x| – \log|C_1| \)
\( -\tan(v) = \log|\frac{x}{C_1}| \)
\( \tan(v) = – \log|\frac{x}{C_1}| \)
\( \tan(v) = \log|\frac{C_1}{x}| \)
Finally, substitute back \( v = \frac{y}{x} \):
\( \tan(\frac{y}{x}) = \log|\frac{C_1}{x}| \)
This is the general solution. Using the property of logarithms helps simplify the expression.
In simple words: We first rewrite the equation to get `dy/dx` on one side and identify it as a "homogeneous" type. Then we substitute `y` with `vx` to make the equation simpler. After separating the `v` and `x` parts, we integrate each side and finally replace `v` with `y/x` to get the full answer.

🎯 Exam Tip: Remember the basic integral for \( \sec^2(x) \) is \( \tan(x) \). Also, combining logarithmic constants as \( \log|C/x| \) often leads to a cleaner final solution.

Question 7. Solve \( (1 + 3e^{\frac{y}{x}}) dy + 3e^{\frac{y}{x}} (1 – \frac{y}{x}) dx = 0 \), given that \( y = 0 \) when \( x = 1 \).
Answer: The given differential equation is \( (1 + 3e^{\frac{y}{x}}) dy + 3e^{\frac{y}{x}} (1 – \frac{y}{x}) dx = 0 \).
First, let's find \( \frac{dy}{dx} \):
\( (1 + 3e^{\frac{y}{x}}) dy = -3e^{\frac{y}{x}} (1 – \frac{y}{x}) dx \)
\( \frac{dy}{dx} = \frac{-3e^{\frac{y}{x}} (1 – \frac{y}{x})}{1 + 3e^{\frac{y}{x}}} \)
This is a homogeneous differential equation.
Let's substitute \( y = vx \).
Then, \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into the equation:
\( v + x \frac{dv}{dx} = \frac{-3e^v (1 – v)}{1 + 3e^v} \)
Now, isolate \( x \frac{dv}{dx} \):
\( x \frac{dv}{dx} = \frac{-3e^v (1 – v)}{1 + 3e^v} - v \)
\( x \frac{dv}{dx} = \frac{-3e^v + 3ve^v - v(1 + 3e^v)}{1 + 3e^v} \)
\( x \frac{dv}{dx} = \frac{-3e^v + 3ve^v - v - 3ve^v}{1 + 3e^v} \)
\( x \frac{dv}{dx} = \frac{-3e^v - v}{1 + 3e^v} \)
\( x \frac{dv}{dx} = - \frac{v + 3e^v}{1 + 3e^v} \)
Separate the variables:
\( \frac{1 + 3e^v}{v + 3e^v} dv = - \frac{dx}{x} \)
Integrate both sides:
\( \int \frac{1 + 3e^v}{v + 3e^v} dv = - \int \frac{dx}{x} \)
Notice that the derivative of the denominator \( (v + 3e^v) \) is \( (1 + 3e^v) \).
So, \( \log|v + 3e^v| = - \log|x| + \log|C| \)
\( \log|v + 3e^v| = \log|\frac{C}{x}| \)
\( v + 3e^v = \frac{C}{x} \)
Substitute back \( v = \frac{y}{x} \):
\( \frac{y}{x} + 3e^{\frac{y}{x}} = \frac{C}{x} \)
Multiply by \( x \):
\( y + 3xe^{\frac{y}{x}} = C \)
Now, use the given condition: \( y = 0 \) when \( x = 1 \).
\( 0 + 3(1)e^{\frac{0}{1}} = C \)
\( 3e^0 = C \)
\( 3(1) = C \)
\( C = 3 \)
Substitute \( C = 3 \) back into the general solution:
\( y + 3xe^{\frac{y}{x}} = 3 \)
This is the particular solution. Applying initial conditions helps find the specific constant.
In simple words: We first rewrite the equation to find `dy/dx` and see it's a "homogeneous" type. Then, we substitute `y` with `vx` to simplify the problem. After separating the `v` and `x` parts, we integrate them. We use the given conditions (`y=0` when `x=1`) to find the exact value of the constant in our solution, giving us the final specific answer.

🎯 Exam Tip: After finding the general solution, remember to use the given initial conditions to find the specific value of the integration constant. This is crucial for solving "initial value problems."

Question 8. Solve \( (x² + y²) dy = xy dx \). It is given that \( y (1) = y(x_0) = e \). Find the value of \( x_0 \).
Answer: The given differential equation is \( (x² + y²) dy = xy dx \).
First, let's express \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{xy}{x² + y²} \)
This is a homogeneous differential equation.
Let's substitute \( y = vx \).
Then, \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into the equation:
\( v + x \frac{dv}{dx} = \frac{x(vx)}{x² + (vx)²} \)
\( v + x \frac{dv}{dx} = \frac{vx²}{x² + v²x²} \)
\( v + x \frac{dv}{dx} = \frac{vx²}{x²(1 + v²)} \)
\( v + x \frac{dv}{dx} = \frac{v}{1 + v²} \)
Now, isolate \( x \frac{dv}{dx} \):
\( x \frac{dv}{dx} = \frac{v}{1 + v²} - v \)
\( x \frac{dv}{dx} = \frac{v - v(1 + v²)}{1 + v²} \)
\( x \frac{dv}{dx} = \frac{v - v - v³}{1 + v²} \)
\( x \frac{dv}{dx} = \frac{-v³}{1 + v²} \)
Separate the variables:
\( \frac{1 + v²}{-v³} dv = \frac{dx}{x} \)
\( - (\frac{1}{v³} + \frac{v²}{v³}) dv = \frac{dx}{x} \)
\( - (v^{-3} + v^{-1}) dv = \frac{dx}{x} \)
Integrate both sides:
\( - \int (v^{-3} + v^{-1}) dv = \int \frac{dx}{x} \)
\( - (\frac{v^{-2}}{-2} + \log|v|) = \log|x| + C \)
\( \frac{1}{2v²} - \log|v| = \log|x| + C \)
Rearrange terms:
\( \frac{1}{2v²} = \log|x| + \log|v| + C \)
\( \frac{1}{2v²} = \log|xv| + C \)
Substitute back \( v = \frac{y}{x} \):
\( \frac{1}{2(\frac{y}{x})²} = \log|x(\frac{y}{x})| + C \)
\( \frac{x²}{2y²} = \log|y| + C \)
Now, use the initial condition \( y(1) = 1 \) (meaning \( y=1 \) when \( x=1 \)) to find \( C \). The question states \( y(1)=e \), so we use that.
Using \( y=e \) when \( x=1 \):
\( \frac{1²}{2e²} = \log|e| + C \)
\( \frac{1}{2e²} = 1 + C \)
\( C = \frac{1}{2e²} - 1 \)
Now, the particular solution is:
\( \frac{x²}{2y²} = \log|y| + \frac{1}{2e²} - 1 \)
We are also given \( y(x_0) = e \), meaning \( y = e \) when \( x = x_0 \). Substitute these values into the particular solution:
\( \frac{x_0²}{2e²} = \log|e| + \frac{1}{2e²} - 1 \)
\( \frac{x_0²}{2e²} = 1 + \frac{1}{2e²} - 1 \)
\( \frac{x_0²}{2e²} = \frac{1}{2e²} \)
Multiply both sides by \( 2e² \):
\( x_0² = 1 \)
\( x_0 = \pm 1 \)
Since the initial condition was \( y(1)=e \), \( x_0 \) could be 1 or -1. The context usually implies a positive value, but mathematically both are valid. This shows how initial conditions help narrow down the possibilities.
In simple words: We first change the equation to find `dy/dx` and notice it is "homogeneous." We replace `y` with `vx` to simplify it. After separating and integrating, we use the first given condition (`y=e` when `x=1`) to find the constant in the solution. Then, we use the second condition (`y=e` when `x=x₀`) with the found constant to solve for `x₀`.

🎯 Exam Tip: When given multiple conditions, use the first condition to find the constant of integration, then use the second condition with the specific constant to solve for the unknown variable. Pay attention to whether the domain allows for positive and negative values.

Question 8. (x² + y²) dy = xy dx. It is given that y (1) = y(x₀) = e. Find the value of x₀.
Answer: The given differential equation can be rewritten to separate \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{xy}{x^2 + y^2} \)
This is a homogeneous differential equation because all terms have the same degree (degree 2). To solve it, we substitute \( y = vx \).
So, \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into the equation:
\( v + x \frac{dv}{dx} = \frac{x(vx)}{x^2 + (vx)^2} \)
\( v + x \frac{dv}{dx} = \frac{vx^2}{x^2 + v^2x^2} \)
\( v + x \frac{dv}{dx} = \frac{vx^2}{x^2(1 + v^2)} \)
\( v + x \frac{dv}{dx} = \frac{v}{1 + v^2} \)
Now, we separate the variables:
\( x \frac{dv}{dx} = \frac{v}{1 + v^2} - v \)
\( x \frac{dv}{dx} = \frac{v - v(1 + v^2)}{1 + v^2} \)
\( x \frac{dv}{dx} = \frac{v - v - v^3}{1 + v^2} \)
\( x \frac{dv}{dx} = \frac{-v^3}{1 + v^2} \)
\( \frac{1 + v^2}{v^3} dv = -\frac{1}{x} dx \)
Next, integrate both sides. We can split the left side:
\( \int \left(\frac{1}{v^3} + \frac{v^2}{v^3}\right) dv = \int -\frac{1}{x} dx \)
\( \int \left(v^{-3} + \frac{1}{v}\right) dv = \int -\frac{1}{x} dx \)
\( -\frac{1}{2}v^{-2} + \log|v| = -\log|x| + \log|C| \)
\( \log|v| - \frac{1}{2v^2} = \log\left|\frac{C}{x}\right| \)
Now, we rearrange the terms and substitute back \( v = \frac{y}{x} \):
\( \log\left|\frac{y}{x}\right| - \frac{1}{2\left(\frac{y}{x}\right)^2} = \log\left|\frac{C}{x}\right| \)
\( \log\left|\frac{y}{x}\right| - \frac{x^2}{2y^2} = \log\left|\frac{C}{x}\right| \)
\( \log\left|\frac{y}{x}\right| - \log\left|\frac{C}{x}\right| = \frac{x^2}{2y^2} \)
\( \log\left|\frac{y/x}{C/x}\right| = \frac{x^2}{2y^2} \)
\( \log\left|\frac{y}{C}\right| = \frac{x^2}{2y^2} \)
This means that:
\( \frac{y}{C} = e^{\frac{x^2}{2y^2}} \)
\( y = C e^{\frac{x^2}{2y^2}} \)

Now we use the given conditions to find the value of \( x_0 \).
First condition: \( y(1) = 1 \). This means when \( x = 1 \), \( y = 1 \).
\( 1 = C e^{\frac{1^2}{2(1)^2}} \)
\( 1 = C e^{\frac{1}{2}} \)
\( C = \frac{1}{e^{1/2}} = \frac{1}{\sqrt{e}} \)
So, the particular solution becomes:
\( y = \frac{1}{\sqrt{e}} e^{\frac{x^2}{2y^2}} \)

Second condition: \( y(x_0) = e \). This means when \( x = x_0 \), \( y = e \).
\( e = \frac{1}{\sqrt{e}} e^{\frac{x_0^2}{2e^2}} \)
Multiply both sides by \( \sqrt{e} \):
\( e \cdot \sqrt{e} = e^{\frac{x_0^2}{2e^2}} \)
\( e^1 \cdot e^{1/2} = e^{\frac{x_0^2}{2e^2}} \)
\( e^{1 + 1/2} = e^{\frac{x_0^2}{2e^2}} \)
\( e^{3/2} = e^{\frac{x_0^2}{2e^2}} \)
Since the bases are equal, the exponents must also be equal:
\( \frac{3}{2} = \frac{x_0^2}{2e^2} \)
Multiply both sides by \( 2e^2 \):
\( 3e^2 = x_0^2 \)
Take the square root of both sides to find \( x_0 \):
\( x_0 = \pm \sqrt{3e^2} \)
\( x_0 = \pm e\sqrt{3} \)
In simple words: We solved the given equation by changing the variables, then put the known values of x and y into the solution. This helped us find the special number, \( x_0 \), that makes the equation true for the second set of conditions. We simplify the powers and then solve for \( x_0 \).

🎯 Exam Tip: Remember to apply both initial conditions carefully. The first condition helps you find the constant of integration, making the general solution specific, and the second condition uses this specific solution to find the required value.