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ডিফারেনশিয়াল ইকুয়েশন সমাধান: সম্পূর্ণ নোট

এই নোটে একটি প্রথম-অর্ডার ডিফারেনশিয়াল ইকুয়েশন ধাপে ধাপে সমাধানের উদাহরণ এবং সাধারণ পদ্ধতি ও কৌশল আলোচনা করা হয়েছে।

উদাহরণ: একটি ডিফারেনশিয়াল ইকুয়েশনের সমাধান

ধাপ ১: Exact Equation কি না তা পরীক্ষা করা 🧐

ধরি একটি সমীকরণ আছে:

\[ M(x,y)\,dx + N(x,y)\,dy = 0 \]

এটি exact হবে যদি:

\[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \]

আমাদের দেওয়া সমীকরণটি:

\[ M(x,y) = 12y + 4y^3 + 6x^2, \quad N(x,y) = 3x + 3xy^2 \]

ডেরিভেটিভগুলো হলো:

\[ \frac{\partial M}{\partial y} = 12 + 12y^2, \quad \frac{\partial N}{\partial x} = 3 + 3y^2 \]

যেহেতু

\[ \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}, \]

সুতরাং সমীকরণটি Exact নয়।

ধাপ ২: Integrating Factor (I.F.) বের করা ⚙️

যদি সমীকরণ Exact না হয়, তবে Integrating Factor (I.F.) খুঁজতে হবে।

N দিয়ে ভাগ করার সূত্র:

\[ f(x) = \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{(12+12y^2) - (3+3y^2)}{3x + 3xy^2} = \frac{9+9y^2}{3x(1+y^2)} = \frac{3}{x} \]

যেহেতু এটি শুধুমাত্র \(x\)-এর ফাংশন, তাই I.F. হবে:

\[ I.F. = e^{\int f(x)\,dx} = e^{\int \frac{3}{x}\,dx} = e^{3 \ln x} = x^3 \]

ধাপ ৩: Exact Equation-এ রূপান্তর ✅

মূল সমীকরণকে \(I.F. = x^3\) দিয়ে গুণ করি:

\[ (12y + 4y^3 + 6x^2)x^3 \, dx + (3x + 3xy^2)x^3 \, dy = 0 \]

নতুন \(M_1\) এবং \(N_1\) হলো:

\[ M_1(x,y) = 12x^3y + 4x^3y^3 + 6x^5, \quad N_1(x,y) = 3x^4 + 3x^4y^2 \]

এখন:

\[ \frac{\partial M_1}{\partial y} = 12x^3 + 12x^3y^2, \quad \frac{\partial N_1}{\partial x} = 12x^3 + 12x^3y^2 \]

দুটো সমান → সমীকরণটি এখন Exact।

ধাপ ৪: Exact Equation সমাধান

Potential function \(F(x,y)\) বের করি:

\[ \frac{\partial F}{\partial x} = M_1, \quad \frac{\partial F}{\partial y} = N_1 \]
\[ F(x,y) = \int M_1\,dx = \int (12x^3y + 4x^3y^3 + 6x^5)\,dx = 3x^4y + x^4y^3 + x^6 + h(y) \]

এখন \(y\)-এর সাথে তুলনা করে:

\[ \frac{\partial F}{\partial y} = 3x^4 + 3x^4y^2 + h'(y) = N_1 \]

সুতরাং \(h'(y) = 0 \Rightarrow h(y) = C\)

চূড়ান্ত সমাধান

\[ \boxed{x^6 + x^4y^3 + 3x^4y = C} \]

ডিফারেনশিয়াল ইকুয়েশন সমাধানের সাধারণ ফ্লোচার্ট 🗺️

  1. স্ট্যান্ডার্ড ফর্মে নিন:
\[ M(x,y)\,dx + N(x,y)\,dy = 0 \]
  1. Exactness পরীক্ষা করুন:
\[ \frac{\partial M}{\partial y} \stackrel{!}{=} \frac{\partial N}{\partial x} \]

  1. Non-Exact হলে I.F. বের করুন:

  2. \(f(x) = \dfrac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N}\) (শুধু \(x\)-এর ফাংশন হলে)

  3. অথবা \(g(y) = \dfrac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M}\) (শুধু \(y\)-এর ফাংশন হলে)

  4. I.F. গণনা:

\[ I.F. = e^{\int f(x)\,dx} \quad \text{অথবা} \quad I.F. = e^{\int g(y)\,dy} \]
  1. Exact Equation তৈরি ও সমাধান করুন:
\[ \int M\,dx + \int \big(N - (\text{যেসব পদে } x \text{ আছে})\big)\,dy = C \]

I.F. সূত্র নির্বাচনের কৌশল 💡

  • N দিয়ে ভাগ:
\[ \frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right) \]

শুধু \(x\)-এর ফাংশন হলে ব্যবহার করুন।

  • M দিয়ে ভাগ:
\[ \frac{1}{M}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) \]

শুধু \(y\)-এর ফাংশন হলে ব্যবহার করুন।

উদাহরণ:

\[ y \log y \,dx + (x - \log y) \,dy = 0 \]
  • \(M = y \log y\), \(N = x - \log y\)
  • \(\frac{\partial M}{\partial y} = \log y + 1\), \(\frac{\partial N}{\partial x} = 1\)

N দিয়ে ভাগ:

\[ \frac{1}{N}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right) = \frac{\log y}{x - \log y} \quad (\text{x ও y দুটো আছে}) \; ❌ \]

M দিয়ে ভাগ:

\[ \frac{1}{M}\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right) = -\frac{1}{y} \quad (\text{শুধু y}) \; ✅ \]

সুতরাং, I.F. বের করতে হবে দ্বিতীয় সূত্র ব্যবহার করে।

ধাপ ১: সমীকরণটিকে স্ট্যান্ডার্ড ফর্মে সাজানো

প্রথমে, আমরা সমীকরণটিকে \(M(x,y)dx + N(x,y)dy = 0\) আকারে সাজাবো।

\[ y^2(ydx + 2xdy) - x^2(2ydx + xdy) = 0 \]
\[ y^3dx + 2xy^2dy - 2x^2ydx - x^3dy = 0 \]

এবার \(dx\) এবং \(dy\) এর সহগগুলো একত্রিত করি:

\[ (y^3 - 2x^2y)\,dx + (2xy^2 - x^3)\,dy = 0 \]

এখানে,

  • \(M(x,y) = y^3 - 2x^2y\)
  • \(N(x,y) = 2xy^2 - x^3\)

ধাপ ২: Exactness পরীক্ষা করা

এখন সমীকরণটি exact কি না, তা পরীক্ষা করতে হবে।

\[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(y^3 - 2x^2y) = 3y^2 - 2x^2 \]
\[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(2xy^2 - x^3) = 2y^2 - 3x^2 \]

যেহেতু \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), সমীকরণটি exact নয়।

ধাপ ৩: Integrating Factor (I.F.) নির্ণয়

যেহেতু সমীকরণটি exact নয়, আমাদের একটি Integrating Factor (I.F.) বের করতে হবে।

আগের গণিতের মতো, আমরা প্রথমে দুটি সাধারণ নিয়ম চেষ্টা করবো:

1.

\[ \frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right) = \frac{(3y^2 - 2x^2) - (2y^2 - 3x^2)}{2xy^2 - x^3} = \frac{x^2+y^2}{x(2y^2-x^2)} \]

→ এটি শুধু \(x\)-এর ফাংশন নয়।

2.

\[ \frac{1}{M}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) = \frac{(2y^2 - 3x^2) - (3y^2 - 2x^2)}{y^3 - 2x^2y} = \frac{-(x^2+y^2)}{y(y^2-2x^2)} \]

→ এটি শুধু \(y\)-এর ফাংশন নয়।

যেহেতু সাধারণ নিয়মগুলো কাজ করছে না, আমাদের অন্য পদ্ধতি দেখতে হবে।
এই সমীকরণটি একটি homogeneous সমীকরণ, তাই এর I.F. \(x^a y^b\) আকারের হবে।
\(a\) এবং \(b\) এর মান বের করার পর দেখা যায়:

\[ I.F. = xy \]

ধাপ ৪: সমীকরণকে Exact বানানো

এখন মূল সমীকরণকে \(I.F. = xy\) দিয়ে গুণ করে নতুন একটি exact সমীকরণ তৈরি করি।

\[ xy(y^3 - 2x^2y)\,dx + xy(2xy^2 - x^3)\,dy = 0 \]
\[ (xy^4 - 2x^3y^2)\,dx + (2x^2y^3 - x^4y)\,dy = 0 \]

এই নতুন সমীকরণটি এখন exact।

ধাপ ৫: Exact সমীকরণের সমাধান

এখন আমরা exact সমীকরণটি সমাধান করবো।

  • নতুন \(M_{new} = xy^4 - 2x^3y^2\)
  • নতুন \(N_{new} = 2x^2y^3 - x^4y\)

সমাধানের সূত্রটি হলো:

\[ \int M_{new}\,dx + \int (\text{N-এর x মুক্ত পদ})\,dy = C \]

এখানে \(N_{new}\)-এর প্রতিটি পদে \(x\) আছে, তাই দ্বিতীয় অংশটি ০ হবে।

\[ \int (xy^4 - 2x^3y^2)\,dx = C \]

\(y\)-কে ধ্রুবক ধরে ইন্টিগ্রেশন করি:

\[ \frac{x^2}{2}y^4 - \frac{x^4}{2}y^2 = C \]

অথবা,

\[ x^2y^4 - x^4y^2 = C_1 \]
\[ x^2y^2(y^2 - x^2) = C_1 \]

এটিই হলো নির্ণেয় সমাধান।

\((12y + 4y^3 + 6x^2)dx + 3(x + xy^2)dy = 0\)

ধাপ ১: স্ট্যান্ডার্ড ফর্ম

প্রদত্ত সমীকরণ:

\[ (12y + 4y^3 + 6x^2)dx + 3(x + xy^2)dy = 0 \]

যেখানে

\[ M(x,y) = 12y + 4y^3 + 6x^2, \quad N(x,y) = 3x + 3xy^2 \]

ধাপ ২: Exactness পরীক্ষা

\[ \frac{\partial M}{\partial y} = 12 + 12y^2, \quad \frac{\partial N}{\partial x} = 3 + 3y^2 \]

যেহেতু \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), সমীকরণটি exact নয়।

ধাপ ৩: Integrating Factor (I.F.) নির্ণয়

ফর্মুলা:

\[ f(x) = \frac{1}{N} \Big(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\Big) \]

প্রদত্ত সমীকরণের জন্য:

\[ f(x) = \frac{(12 + 12y^2) - (3 + 3y^2)}{3x + 3xy^2} = \frac{9 + 9y^2}{3x(1 + y^2)} = \frac{3}{x} \]

যেহেতু \(f(x)\) কেবল \(x\)-এর উপর নির্ভর করে, I.F. হবে:

\[ I.F. = e^{\int f(x) dx} = e^{\int \frac{3}{x} dx} = e^{3 \ln x} = x^3 \]
  • I.F. = \(x^3\)

ধাপ ৪: সমীকরণকে Exact বানানো

মূল সমীকরণে I.F. দ্বারা গুণ করলে:

\[ x^3(12y + 4y^3 + 6x^2)dx + x^3(3x + 3xy^2)dy = 0 \]
\[ (12x^3y + 4x^3y^3 + 6x^5)dx + (3x^4 + 3x^4y^2)dy = 0 \]

ধাপ ৫: Exact সমীকরণের সমাধান

নতুন exact সমীকরণে:

\[ M_{new} = 12x^3y + 4x^3y^3 + 6x^5, \quad N_{new} = 3x^4 + 3x^4y^2 \]

সমাধান সূত্র:

\[ \int M_{new}\,dx + \int (\text{N-এর x মুক্ত পদ})\,dy = C \]

\(N_{new}\)-এর প্রতিটি পদে \(x\) থাকায় দ্বিতীয় অংশটি ০ হবে।

\[ \int (12x^3y + 4x^3y^3 + 6x^5)dx = C \]
\[ 12y\frac{x^4}{4} + 4y^3\frac{x^4}{4} + 6\frac{x^6}{6} = C \]
\[ 3x^4y + x^4y^3 + x^6 = C \]

সাজিয়ে লিখলে:

\[ x^6 + x^4y^3 + 3x^4y = C \]

এটিই হলো নির্ণেয় সমাধান।

Example: Solve the differential equation

Step 1: Standard Form

Write the equation in standard form \(M(x,y)dx + N(x,y)dy = 0\):

\[ (12y + 4y^3 + 6x^2)dx + 3(x + xy^2)dy = 0 \]

Here,

\[ M(x,y) = 12y + 4y^3 + 6x^2, \quad N(x,y) = 3x + 3xy^2 \]

Step 2: Test for Exactness

Compute the partial derivatives:

\[ \frac{\partial M}{\partial y} = 12 + 12y^2, \quad \frac{\partial N}{\partial x} = 3 + 3y^2 \]

Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), the equation is not exact.

Step 3: Find the Integrating Factor (I.F.)

Use the formula:

\[ f(x) = \frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right) \]

Substitute the values:

\[ f(x) = \frac{(12 + 12y^2) - (3 + 3y^2)}{3x + 3xy^2} = \frac{3}{x} \]

Since \(f(x)\) depends only on \(x\), the Integrating Factor is:

\[ I.F. = e^{\int f(x) dx} = e^{3 \ln x} = x^3 \]

Step 4: Make the Equation Exact

Multiply the original equation by \(x^3\):

\[ x^3(12y + 4y^3 + 6x^2)dx + x^3(3x + 3xy^2)dy = 0 \]

Simplify:

\[ (12x^3y + 4x^3y^3 + 6x^5)dx + (3x^4 + 3x^4y^2)dy = 0 \]

Step 5: Solve the Exact Equation

Here,

\[ M_{new} = 12x^3y + 4x^3y^3 + 6x^5, \quad N_{new} = 3x^4 + 3x^4y^2 \]

Since all terms in \(N_{new}\) contain \(x\), the second integral is \(0\). Integrate \(M_{new}\) with respect to \(x\), treating \(y\) as constant:

\[ \int (12x^3y + 4x^3y^3 + 6x^5) dx = 3x^4y + x^4y^3 + x^6 = C \]

Final solution:

\[ x^6 + x^4y^3 + 3x^4y = C \]

\(y \log y \,dx + (x - \log y) \,dy = 0\)

Step 1: Standard Form

The equation is already in the standard form \(M(x,y)dx + N(x,y)dy = 0\).

\[y \log y \,dx + (x - \log y) \,dy = 0\]

Here, we can identify:
* \(M(x,y) = y \log y\)
* \(N(x,y) = x - \log y\)

Step 2: Test for Exactness

Next, we check if the equation is exact by calculating the partial derivatives.

  • To find \(\frac{\partial M}{\partial y}\), we use the product rule:

    \[\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(y \log y) = (1)(\log y) + y\left(\frac{1}{y}\right) = \log y + 1\]

  • Now we find \(\frac{\partial N}{\partial x}\):

    \[\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x - \log y) = 1\]

Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), the equation is not exact.

Step 3: Find the Integrating Factor (I.F.)

Since the equation is not exact, we need an Integrating Factor (I.F.).

First, we test the rule for a function of \(x\):

\[\frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right) = \frac{(\log y + 1) - 1}{x - \log y} = \frac{\log y}{x - \log y}\]

This expression contains both \(x\) and \(y\), so it won't work.

Next, we test the rule for a function of \(y\):

\[g(y) = \frac{1}{M}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)\]
\[g(y) = \frac{1 - (\log y + 1)}{y \log y} = \frac{1 - \log y - 1}{y \log y} = \frac{-\log y}{y \log y} = -\frac{1}{y}\]

This result is a function of \(y\) only, so we can use it to find the I.F.

\[I.F. = e^{\int g(y)dy} = e^{\int -\frac{1}{y}dy} = e^{-\ln y} = e^{\ln(y^{-1})} = y^{-1} = \frac{1}{y}\]

The Integrating Factor is \(\frac{1}{y}\).

Step 4: Make the Equation Exact

Now, we multiply the original equation by the I.F. (\(\frac{1}{y}\)) to make it exact.

\[\frac{1}{y} \big(y \log y \,dx\big) + \frac{1}{y} \big((x - \log y) \,dy\big) = 0\]
\[\log y \,dx + \left(\frac{x}{y} - \frac{\log y}{y}\right)dy = 0\]

Step 5: Solve the Exact Equation

This new equation is exact. The solution is found using the formula:

\[ \int\limits_{\substack{y=\text{const.}}} M_{new}\,dx \;+\; \int (\text{terms in } N_{new} \text{ free from } x)\,dy = C \]

Here,

  • \(M_{new} = \log y\)
  • \(N_{new} = \frac{x}{y} - \frac{\log y}{y}\)

The term in \(N_{new}\) that is free from \(x\) is \(-\frac{\log y}{y}\).

Now we integrate:

\[\int \log y \,dx + \int \left(-\frac{\log y}{y}\right)dy = C\]

  • First integral (treating \(y\) as a constant):

    \[\int \log y \,dx = x \log y\]

  • Second integral (using substitution, let \(u = \log y\)):

    \[ \int \left(-\frac{\log y}{y}\right)dy = -\int u \,du = -\frac{u^2}{2} = -\frac{(\log y)^2}{2} \]

Combining the parts gives the final solution:

\[x \log y - \frac{(\log y)^2}{2} = C\]

Solution Method:01: \(y^2(ydx + 2xdy) - x^2(2ydx + xdy) = 0\)

Step 1: Standard Form

First, we need to expand the equation and arrange it into the standard form \(M(x,y)dx + N(x,y)dy = 0\).

\[y^2(ydx + 2xdy) - x^2(2ydx + xdy) = 0$$ $$y^3dx + 2xy^2dy - 2x^2ydx - x^3dy = 0\]

Now, we group the \(dx\) and \(dy\) terms:

\[(y^3 - 2x^2y)dx + (2xy^2 - x^3)dy = 0\]

From this, we can identify:
* \(M(x,y) = y^3 - 2x^2y\)
* \(N(x,y) = 2xy^2 - x^3\)

Step 2: Test for Exactness

Next, we check if the equation is exact by calculating the partial derivatives.

\[\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(y^3 - 2x^2y) = 3y^2 - 2x^2\]
\[\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(2xy^2 - x^3) = 2y^2 - 3x^2\]

Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), the equation is not exact.

Step 3: Find the Integrating Factor (I.F.)

We need to find an Integrating Factor (I.F.). The two standard rules for finding an I.F. do not yield a function of a single variable in this case.

However, since the equation is a homogeneous differential equation, we can look for an integrating factor of the form \(x^a y^b\). By solving for the exponents, we find that \(a=1\) and \(b=1\).

Therefore, the Integrating Factor is \(xy\).

Step 4: Make the Equation Exact

Now, we multiply the standard form of the equation by the I.F. (\(xy\)) to make it exact.

\[xy \big((y^3 - 2x^2y)dx\big) + xy \big((2xy^2 - x^3)dy\big) = 0\]
\[(xy^4 - 2x^3y^2)dx + (2x^2y^3 - x^4y)dy = 0\]

Step 5: Solve the Exact Equation

This new equation is exact. The solution is found using the formula:

\[\int M_{new}\,dx + \int (\text{terms in } N_{new} \text{ free from } x)\,dy = C\]

Here,
* \(M_{new} = xy^4 - 2x^3y^2\)
* \(N_{new} = 2x^2y^3 - x^4y\)

Since all terms in \(N_{new}\) contain \(x\), the second integral is \(0\). We only need to integrate \(M_{new}\) with respect to \(x\), treating \(y\) as a constant.

\[\int (xy^4 - 2x^3y^2)dx = C\]
\[y^4\left(\frac{x^2}{2}\right) - 2y^2\left(\frac{x^4}{4}\right) = C\]
\[\frac{x^2y^4}{2} - \frac{x^4y^2}{2} = C\]

To simplify, we can multiply the entire equation by 2 (and absorb it into the constant \(C\)):

\[x^2y^4 - x^4y^2 = C_1\]

Factoring out the common terms gives the final solution:

\[x^2y^2(y^2 - x^2) = C_1\]

Solution Method:02 \(y^2(ydx + 2xdy) - x^2(2ydx + xdy) = 0\)

Step 1: Rearrange into \(\frac{dy}{dx}\) form

First, we expand and group the terms to isolate \(\frac{dy}{dx}\).

\[y^3dx + 2xy^2dy - 2x^2ydx - x^3dy = 0\]
\[(2xy^2 - x^3)dy = (2x^2y - y^3)dx\]
\[\frac{dy}{dx} = \frac{2x^2y - y^3}{2xy^2 - x^3}\]

This is a homogeneous equation because all terms in the numerator and denominator are of the same degree (degree 3).

Step 2: Apply the Substitution

We use the substitution \(y=vx\), which implies \(\frac{dy}{dx} = v + x\frac{dv}{dx}\).

Substitute these into the equation:

\[v + x\frac{dv}{dx} = \frac{2x^2(vx) - (vx)^3}{2x(vx)^2 - x^3}\]

Now, we simplify the right side by factoring out \(x^3\):

\[v + x\frac{dv}{dx} = \frac{x^3(2v - v^3)}{x^3(2v^2 - 1)} = \frac{2v - v^3}{2v^2 - 1}\]

Step 3: Separate the Variables

Next, we isolate the \(v\) and \(x\) variables.

\[x\frac{dv}{dx} = \frac{2v - v^3}{2v^2 - 1} - v\]
\[x\frac{dv}{dx} = \frac{2v - v^3 - v(2v^2 - 1)}{2v^2 - 1}\]
\[x\frac{dv}{dx} = \frac{2v - v^3 - 2v^3 + v}{2v^2 - 1} = \frac{3v - 3v^3}{2v^2 - 1}\]
\[x\frac{dv}{dx} = \frac{3v(1 - v^2)}{2v^2 - 1}\]

Now, we rearrange to separate the variables for integration:

\[\frac{2v^2 - 1}{3v(1 - v^2)}dv = \frac{1}{x}dx\]

Step 4: Integrate Both Sides

The left side requires integration by partial fractions, which breaks it down into:

\[\frac{1}{3} \int \left( -\frac{1}{v} + \frac{1/2}{1-v} - \frac{1/2}{1+v} \right)dv = \int \frac{1}{x}dx\]

Integrating both sides gives us:

\[\frac{1}{3} \left[ -\ln|v| - \frac{1}{2}\ln|1-v| - \frac{1}{2}\ln|1+v| \right] = \ln|x| + C\]

Let's simplify the logarithmic terms:

\[-\ln|v| - \frac{1}{2}\ln|1-v^2| = 3\ln|x| + C_1\]
\[\ln|x^3| + \ln|v| + \ln|\sqrt{1-v^2}| = C_2\]
\[\ln|x^3 v \sqrt{1-v^2}| = C_2\]
\[x^3 v \sqrt{1-v^2} = C_3\]

Step 5: Substitute Back

Finally, substitute \(v = \frac{y}{x}\) back into the equation.

\[x^3 \left(\frac{y}{x}\right) \sqrt{1 - \left(\frac{y}{x}\right)^2} = C_3\]
\[x^2 y \sqrt{\frac{x^2 - y^2}{x^2}} = C_3\]
\[x^2 y \frac{\sqrt{x^2 - y^2}}{x} = C_3\]
\[xy \sqrt{x^2 - y^2} = C_3\]

Squaring both sides gives the final, simplified solution:

\[x^2y^2(x^2 - y^2) = C\]

$\((1 + xy)y \,dx + (1 - xy)x \,dy = 0\)$

\[(1 + xy)y \,dx + (1 - xy)x \,dy = 0\]
\[y \,dx + xy^2 \,dx + x \,dy - x^2y \,dy = 0\]
\[(y \,dx + x \,dy) = x^2y \,dy - xy^2 \,dx\]
\[d(xy) = xy(x \,dy - y \,dx)\]

Substitute \(x \,dy - y \,dx = x^2 d\left(\frac{y}{x}\right)\)

\[d(xy) = xy \left[x^2 d\left(\frac{y}{x}\right)\right]\]
\[d(xy) = x^3y \,d\left(\frac{y}{x}\right)\]

Rewrite the coefficient \(x^3y = \frac{(xy)^2}{y/x}\)

\[d(xy) = \frac{(xy)^2}{y/x} \,d\left(\frac{y}{x}\right)\]
\[\frac{d(xy)}{(xy)^2} = \frac{d(y/x)}{y/x}\]
\[\int \frac{d(xy)}{(xy)^2} = \int \frac{d(y/x)}{y/x}\]
\[-\frac{1}{xy} = \ln\left|\frac{y}{x}\right| + C\]

s

Given Differential Equation:

\[(6x - 5y + 4)dy + (y - 2x - 1)dx = 0\]

Step 1: Rearrange and Prepare for Substitution

First, we write the equation in the form \(\frac{dy}{dx}\):

\[\frac{dy}{dx} = - \frac{y - 2x - 1}{6x - 5y + 4} = \frac{2x - y + 1}{6x - 5y + 4} \quad \cdots (1)\]

Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) (i.e., \(\frac{2}{6} \neq \frac{-1}{-5}\)), we use the substitution:

Let \(x = x' + h \implies dx = dx'\)
Let \(y = y' + k \implies dy = dy'\)

Equation (1) becomes:

\[\frac{dy'}{dx'} = \frac{2(x' + h) - (y' + k) + 1}{6(x' + h) - 5(y' + k) + 4} = \frac{2x' - y' + (2h - k + 1)}{6x' - 5y' + (6h - 5k + 4)}\]

To make this homogeneous, we solve:

  1. \(2h - k + 1 = 0\)
  2. \(6h - 5k + 4 = 0\)

Correction 1: The values for \(h\) and \(k\) must be calculated correctly.
From (1), \(k = 2h + 1\). Substituting into (2):
\(6h - 5(2h+1) + 4 = 0 \implies 6h - 10h - 5 + 4 = 0 \implies -4h - 1 = 0 \implies h = -\frac{1}{4}\).
Then, \(k = 2(-\frac{1}{4}) + 1 = \frac{1}{2}\).
(Your work had an error in the sign of \(h\)).

Step 2: Solve the Homogeneous Equation

The equation reduces to the homogeneous form:

\[\frac{dy'}{dx'} = \frac{2x' - y'}{6x' - 5y'} \quad \cdots (2)\]

Again, let \(y' = vx'\). Differentiating with respect to \(x'\) gives \(\frac{dy'}{dx'} = v + x' \frac{dv}{dx'} \quad \cdots (3)\).

From (2) and (3):

\[v + x' \frac{dv}{dx'} = \frac{2x' - vx'}{6x' - 5vx'} = \frac{2 - v}{6 - 5v}\]
\[x' \frac{dv}{dx'} = \frac{2 - v}{6 - 5v} - v\]

Correction 2: The next algebraic step in your work contained a sign error.

\[x' \frac{dv}{dx'} = \frac{2 - v - v(6 - 5v)}{6 - 5v} = \frac{2 - v - 6v + 5v^2}{6 - 5v}\]
\[x' \frac{dv}{dx'} = \frac{5v^2 - 7v + 2}{6 - 5v}\]

Step 3: Separate Variables and Integrate

Now, we separate the variables. This will look different from your version due to the corrected algebra above.

\[\frac{dx'}{x'} = \frac{6 - 5v}{5v^2 - 7v + 2} dv$$The denominator factors to $(v-1)(5v-2)$. Using partial fractions:$$\frac{6 - 5v}{(v-1)(5v-2)} = \frac{A}{v-1} + \frac{B}{5v-2}\]

Solving this gives \(A = \frac{1}{3}\) and \(B = -\frac{20}{3}\).

Now, we arrange for integration:

\[\frac{dx'}{x'} = \left[ \frac{1}{3(v-1)} - \frac{20}{3(5v-2)} \right] dv\]
\[\int \frac{dx'}{x'} - \int \frac{1}{3(v-1)} dv + \int \frac{20}{3(5v-2)} dv = 0\]

Now integrating all terms:
$\(\ln|x'| - \frac{1}{3}\ln|v-1| + \frac{20}{3} \cdot \frac{1}{5}\ln|5v-2| = C_1\)$

\[\ln|x'| - \frac{1}{3}\ln|v-1| + \frac{4}{3}\ln|5v-2| = C_1\]

Multiply by 3 to clear the fractions and set \(3C_1 = \ln C\):

\[3\ln|x'| - \ln|v-1| + 4\ln|5v-2| = \ln C$$Using logarithm properties:$$\ln|(x')^3| - \ln|v-1| + \ln|(5v-2)^4| = \ln C\]
\[\ln \left| \frac{(x')^3 (5v-2)^4}{v-1} \right| = \ln C$$$$\frac{(x')^3 (5v-2)^4}{v-1} = C\]
\[(x')^3 (5v-2)^4 = C(v-1)\]

Step 4: Back-Substitute

Substitute \(x' = x + \frac{1}{4}\), \(y' = y - \frac{1}{2}\), and \(v = \frac{y'}{x'} = \frac{y-1/2}{x+1/4} = \frac{4y-2}{4x+1}\).

After substitution and simplification (as shown in the previous detailed answer), we arrive at the final solution:

\[(4y - 4x - 3) = K(5y - 2x - 3)^4\]

where \(K\) is the final constant.

Of course. Here is the solution, following the exact step-by-step format from the image, but with all the necessary corrections.

Given Equation:
$\(\frac{dy}{dx} = \frac{4x+6y+5}{3y+2x+4} = \frac{2(2x+3y)+5}{(2x+3y)+4}\)$

Let \(2x + 3y = t\)
Differentiating with respect to \(x\):

\[2 + 3\frac{dy}{dx} = \frac{dt}{dx}\]
\[\frac{dy}{dx} = \frac{1}{3}\left(\frac{dt}{dx} - 2\right)\]

Substituting this into the original equation:$\(\frac{1}{3}\left(\frac{dt}{dx} - 2\right) = \frac{2t+5}{t+4}\)$

Now, we solve for \(\frac{dt}{dx}\):

\[\frac{1}{3}\frac{dt}{dx} = \frac{2t+5}{t+4} + \frac{2}{3}\]
\[\frac{1}{3}\frac{dt}{dx} = \frac{3(2t+5) + 2(t+4)}{3(t+4)} = \frac{6t+15+2t+8}{3(t+4)}\]
\[\frac{1}{3}\frac{dt}{dx} = \frac{8t+23}{3(t+4)}\]

Canceling the \(\frac{1}{3}\) on both sides gives:

\[\frac{dt}{dx} = \frac{8t+23}{t+4}\]

Now, we separate the variables and integrate:

\[\int \frac{t+4}{8t+23} dt = \int dx\]

To perform the integration on the left, we manipulate the fraction:

\[\frac{1}{8} \int \frac{8(t+4)}{8t+23} dt = x + C\]
\[\frac{1}{8} \int \frac{8t+32}{8t+23} dt = x + C\]
\[\frac{1}{8} \int \frac{(8t+23) + 9}{8t+23} dt = x + C\]
\[\frac{1}{8} \int \left(1 + \frac{9}{8t+23}\right) dt = x + C\]
\[\frac{1}{8} \left[ \int 1 \ dt + \int \frac{9}{8t+23} dt \right] = x + C\]
\[\frac{1}{8} \left[ t + \frac{9}{8}\ln|8t+23| \right] = x + C\]
\[ Multiplying by 8:\]
\[t + \frac{9}{8}\ln|8t+23| = 8x + 8C\]

Now, substitute back \(t = 2x + 3y\):

\[(2x+3y) + \frac{9}{8}\ln|8(2x+3y)+23| = 8x + 8C\]
\[(2x+3y) + \frac{9}{8}\ln|16x+24y+23| = 8x + K \quad (\text{where } K = 8C)\]

Multiplying by 8 again to clear the fraction:

\[8(2x+3y) + 9\ln|16x+24y+23| = 64x + 8K\]
\[16x + 24y + 9\ln|16x+24y+23| = 64x + K' \quad (\text{where } K' = 8K)\]
\[24y - 48x + 9\ln|16x+24y+23| = K'\]

Finally, dividing by 3 for the simplest form:

\[\mathbf{8y - 16x + 3\ln|16x+24y+23| = K''}\]