Class 02

Example ii:
We are given:

\[ \frac{dy}{dx} = e^x \left( A\sin 2x + B\cos 2x \right) + e^x \left[ A\cos 2x \cdot 2 + B(-\sin 2x) \cdot 2 \right] \]

This simplifies to:

\[ \frac{dy}{dx} = y + e^x \left( 2A\cos 2x - 2B\sin 2x \right) \tag{1} \]

Now, differentiating again with respect to \(x\):

\[ \frac{d^2y}{dx^2} = y' + e^x \left[ 2(-2A\sin 2x) - 2(2B\cos 2x) \right] \]

\[ \quad\quad\quad\; + e^x \left( 2A\cos 2x - 2B\sin 2x \right) + \frac{dy}{dx} \]

From equation (1), substituting \(e^x(2A\cos 2x - 2B\sin 2x) = \frac{dy}{dx} - y\), we get:

\[ \frac{d^2y}{dx^2} = -4y + \left( \frac{dy}{dx} - y \right) + \frac{dy}{dx} \]

Simplifying:

\[ \frac{d^2y}{dx^2} = -4y + 2\frac{dy}{dx} - y \]

\[ \frac{d^2y}{dx^2} + 5y - 2\frac{dy}{dx} = 0 \]

Hence,

\[ \boxed{\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 5y = 0} \]

is the required differential equation.

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Example:
Show that \(Ax^2 + By^2 = 1\) is the solution of

\[ x \left\{ y \frac{d^2 y}{dx^2} + \left( \frac{dy}{dx} \right)^2 \right\} = y \frac{dy}{dx} \]

Solution:
Given:

\[ Ax^2 + By^2 = 1 \]

Differentiating with respect to \(x\):

\[ A \cdot 2x + B \cdot 2y \frac{dy}{dx} = 0 \]

\[ 2Ax + 2By \frac{dy}{dx} = 0 \]

\[ Ax + By \frac{dy}{dx} = 0 \]

\[ Ax = -B y \frac{dy}{dx} \]

\[ -\frac{A}{B} = \frac{y}{x} \cdot \frac{dy}{dx} \]

Differentiating again:

\[ 0 = \frac{y}{x} \cdot \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{x \frac{dy}{dx} - y}{x^2} \]

Multiplying through by \(x^2\):

\[ y \cdot x \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0 \]

\[ xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0 \]

Thus,

\[ x \left\{ y \frac{d^2 y}{dx^2} + \left( \frac{dy}{dx} \right)^2 \right\} = y \frac{dy}{dx} \]

Hence proved. \(\quad \text{(Shown)}\) ✅

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Short Question: Concentric Circle Whose Centre is the Origin

1.
A concentric circle with its centre at the origin has the general equation:

\[ x^2 + y^2 = a^2 \]

Differentiating with respect to \(x\), we get:

\[ x + y \frac{dy}{dx} = 0 \]

which is the required differential equation.

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2. For a general circle (not necessarily concentric), the equation is:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

Here, the Centre is \((-g, -f)\).

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আমরা আগে concept clear করব, তারপর equation derive করব।

1️⃣ Problem-টা আসলে কী বলছে?

আমরা একটা family of circles নিয়ে কথা বলছি।
এই circles গুলো special —

• এগুলো x-axis-কে (horizontal axis) origin (0,0)-এ touch করছে।
• Circle-এর center সবসময় y-axis-এর উপরে থাকবে।

📌 মানে, এগুলো এমন হবে ↓
(একটা circle উপরে, যেটা x-axis-এর উপর শুধু এক পয়েন্টে স্পর্শ করছে, নিচে যেতে পারছে না)

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2️⃣ কেন center y-axis-এ?

কারণ:
যদি circle origin-এ x-axis-কে স্পর্শ করে, symmetry-এর কারণে center অবশ্যই x=0 লাইন (y-axis)-এ হবে।
ধরি center = \((0,a)\), radius = \(a\) (কারণ center থেকে x-axis পর্যন্ত vertical distance = a, যা radius)।

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3️⃣ Circle-এর Equation লিখি

Center (0,a), radius a ⇒ equation হবে:

\[ (x-0)^2 + (y-a)^2 = a^2 \]

সেটা expand করলে:

\[ x^2 + y^2 - 2ay = 0 \]

এখানে a হলো আমাদের parameter (প্রতিটা circle-এর জন্য আলাদা)।

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4️⃣ এখন Derivative নেই

আমরা চাই differential equation, মানে parameter a কে বাদ দিয়ে x, y, dy/dx-এর relation।

Equation:

\[ x^2 + y^2 - 2ay = 0 \]

Differentiating wrt x:

\[ 2x + 2y \frac{dy}{dx} - 2a \frac{dy}{dx} = 0 \]

2 দিয়ে ভাগ করলে:

\[ x + (y-a) y' = 0 \]

এখানে \(y' = \frac{dy}{dx}\)।

5️⃣ a কে eliminate করি

Original equation থেকে:

\[ 2ay = x^2 + y^2 \quad\Rightarrow\quad a = \frac{x^2 + y^2}{2y} \]

এখন \(y-a\) বের করি:

\[ y-a = y - \frac{x^2 + y^2}{2y} = \frac{2y^2 - (x^2 + y^2)}{2y} = \frac{y^2 - x^2}{2y} = -\frac{x^2 - y^2}{2y} \]

6️⃣ Back to derivative equation

\(y-a = -\frac{x^2 - y^2}{2y}\) এটা \(x + (y-a)y' = 0\)-এ বসাই:

\[ x - \frac{x^2 - y^2}{2y} y' = 0 \]

Multiply by \(2y\):

\[ 2xy - (x^2 - y^2) y' = 0 \]

Rearrange:

\[ (x^2 - y^2) y' - 2xy = 0 \]

y' এর জায়গায় \(\frac{dy}{dx}\) লিখলে:

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

Cross-multiply করলে:

\[ \boxed{(x^2 - y^2) dy - 2xy dx = 0} \]

✅ এটাই চেয়েছিল।
বোঝার লজিক:

• Geometry থেকে center নির্ধারণ করো
• Equation লেখো
• Differentiate করে parameter বাদ দাও
• Final differential equation পেয়ে যাও

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Example:

Show that the differential equation of a family of circles touching the \(x\)-axis at the origin is

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

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Solution:

The equation of a family of circles touching the \(x\)-axis at the origin is:

\[ x^2 + y^2 + 2fy + c = 0 \]

Since the circle passes through \((0,0)\),

\[ c = 0 \]

Thus,

\[ x^2 + y^2 + 2fy = 0 \quad \cdots (1) \]

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Step 1: Differentiate w.r.to \(x\)

Differentiating (1):

\[ 2x + 2y \frac{dy}{dx} + 2f \frac{dy}{dx} = 0 \]

Divide by 2:

\[ x + y \frac{dy}{dx} + f \frac{dy}{dx} = 0 \quad \cdots (2) \]

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Step 2: Eliminate \(f\)

From (1):

\[ f = -\frac{x^2 + y^2}{2y} \]

Substitute in (2):

\[ x + y \frac{dy}{dx} - \frac{x^2 + y^2}{2y} \cdot \frac{dy}{dx} = 0 \]

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Step 3: Simplify

Multiply through by \(2y\):

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

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

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

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Step 4: Rearrange

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

Multiply by \(-1\):

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

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✅ Hence proved:

\[ \boxed{(x^2 - y^2) \, dy - 2xy \, dx = 0} \]

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H.W.: Find the differential equation of a family of circles touching the \(y\)-axis.

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H.W.–2:
Show that the differential equation of a family of circles

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

is

\[ (1 + y'^2) y_3 - 3y' y''^2 = 0 \]

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Practice Problem — Find Order & Degree

Given equation:

\[ \sqrt[3]{\left(\frac{d^3y}{dx^3}\right)^4 - 5x\frac{d^2y}{dx^2} + y} = \sqrt[5]{\left(\frac{dy}{dx}\right)^2 + y^2 - x} \]

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Step 1: Order

• The highest order derivative present is \(\frac{d^3y}{dx^3}\).
• Therefore, Order = 3.

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Step 2: Degree

• The cube root applies to the entire left-hand side expression, and the fifth root applies to the entire right-hand side.
• Remove both roots by raising both sides to the 15th power (LCM of 3 and 5):

\[ \left[\left(\frac{d^3y}{dx^3}\right)^4 - 5x\frac{d^2y}{dx^2} + y\right]^5 = \left[\left(\frac{dy}{dx}\right)^2 + y^2 - x\right]^3 \]

• Now, the highest derivative term becomes:

\[ \left(\frac{d^3y}{dx^3}\right)^{4 \times 5} = \left(\frac{d^3y}{dx^3}\right)^{20} \]

• Hence, Degree = 20.

Final Answer:

\[ \boxed{\text{Order} = 3, \quad \text{Degree} = 20} \]

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Differential Equation of First Order and First Degree

The most general form of an ordinary differential equation of the 1st order and 1st degree is:

\[ M\,dx + N\,dy = 0 \]

or equivalently,

\[ N \frac{dy}{dx} + M = 0 \]

where both \(M\) and \(N\) are functions of \(x\) and \(y\), or constants, and do not involve any derivatives.

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Solution Methods:

1. Separation of Variables – Variables are separated so that all \(x\) terms are on one side and all \(y\) terms on the other.
2. Homogeneous Equation – Equation in which \(M\) and \(N\) are homogeneous functions of the same degree.
3. Equation Reducible to Homogeneous Form – Transformed into homogeneous type by substitution.
4. Exact Equation – Condition: \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\).
5. Linear Equation – Follows the form \(\frac{dy}{dx} + P(x)y = Q(x)\).
6. Equation Reducible to Linear Form – Can be rearranged into the linear equation format.

Example 1 — Separation of Variables

Solve:

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

Solution:
Given:

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

Divide through by \((x - 1)(y - 1)\):

\[ \frac{x^2}{x - 1}\, dx + \frac{y^2}{y - 1}\, dy = 0 \]

Rewrite:

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

Integrate both sides:

\[ \int x\, dx + \int 1\, dx + \int \frac{1}{x - 1} dx + \int y\, dy + \int 1\, dy + \int \frac{1}{y - 1} dy = C \]

\[ \frac{x^2}{2} + x + \log|x - 1| + \frac{y^2}{2} + y + \log|y - 1| = C \]

Multiply by 2 (or combine constants):

\[ x^2 + y^2 + 2x + 2y + \log(x - 1) + \log(y - 1) = C \]

where \(C\) is an arbitrary constant.

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Example — 2
Solve:

\[ \frac{dy}{dx} = e^{x+y} + x^2 e^{-y} \]

Solution:

\[ \frac{dy}{dx} = e^x \cdot e^y + x^2 \cdot e^{-y} \]

Actually, write directly:

\[ e^y \frac{dy}{dx} = e^x + x^2 \]

Now integrate both sides with respect to \(x\):

\[ \int e^y \, dy = \int e^x \, dx + \int x^2 \, dx \]

\[ e^y = e^x + \frac{x^3}{3} + C \]

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HW-02

❓H.W.–2:
Show that the differential equation of a family of circles

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

(এটা হল general equation of a circle, যেখানে g, f, c constant — different circles এর জন্য g, f, c এর মান আলাদা হতে পারে)

is

\[ (1 + y_1^{\,2})\,y_3 - 3y_1\,y_2^{\,2} = 0, \]

(আমাদের লক্ষ্য এই form-টা প্রমাণ করা, যেখানে \(y_1, y_2, y_3\) হলো প্রথম, দ্বিতীয় ও তৃতীয় derivative)

where \(y_1=\dfrac{dy}{dx},\; y_2=\dfrac{d^2y}{dx^2},\; y_3=\dfrac{d^3y}{dx^3}\).

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Given family:

\[ x^2+y^2+2gx+2fy+c=0 \quad (g,f,c \text{ constants}) \]

1st derivative → \(x+yy_1+g+fy_1=0\) … (A)
(এখানে দুই পাশে \(d/dx\) করলাম; \(y^2\) এর derivative = \(2y \cdot y_1\), আবার \(2fy\) এর derivative = \(2f \cdot y_1\), তারপর 2 দিয়ে divide করায় এই ফর্ম পেলাম)

2nd derivative → \(1+y_1^2+yy_2+fy_2=0\)

(আবার derivative করলাম; \(y y_1\) এর derivative = \(y_1^2 + y y_2\))

\[ \Rightarrow\ (y+f)\,y_2=-(1+y_1^2) \quad\text{… (B)} \]

(এখানে \(y_2\) এর term গুলো একসাথে করে, \(y+f\) factor হিসেবে বের করে নিলাম)

3rd derivative of (2nd) → \(3y_1y_2+(y+f)\,y_3=0\) … (C)

(এবার দ্বিতীয় derivative সমীকরণটার derivative করলাম; \(y_1^2\) এর derivative = \(2y_1y_2\), সাথে \((y+f) y_2\) এর derivative দিলে \(y_1 y_2 + (y+f) y_3\) — সব মিলে \(3y_1 y_2 + (y+f) y_3\))

(B) থেকে \((y+f)=-(1+y_1^2)/y_2\) বসিয়ে (C):
(আগের ধাপের (B) থেকে \(y+f\) এর মান নিলাম এবং (C)-তে বসালাম)

\[ 3y_1y_2-\frac{1+y_1^2}{y_2}\,y_3=0 \]

(এখানে \(y+f\) এর জায়গায় fraction বসেছে, আর minus sign কে adjust করে এই ফর্ম এসেছে)

\[ \boxed{(1+y_1^2)\,y_3-3y_1\,y_2^2=0} \]

(শেষে rearrange করে ঠিক আমাদের প্রমাণ করতে চাওয়া final equation পেলাম ✅)