🌱Differential Equationsअवकल समीकरणDifferential Equations

1. Only the first derivative appears.केवल प्रथम अवकलज है।Sirf first derivative hai.
   \text{(i) highest derivative } \dfrac{dy}{dx} \Rightarrow \text{order } 1
2. It is a polynomial equation in y' and the highest power of y' is one.यह y' में बहुपद समीकरण है और y' की उच्चतम घात एक है।Ye y' mein polynomial equation hai aur y' ki highest power ek hai.
   \text{(i) power of } \dfrac{dy}{dx} \text{ is } 1 \Rightarrow \text{degree } 1
3. \frac{dy}{dx} also appears, but the second derivative is higher.\frac{dy}{dx} भी है, पर द्वितीय अवकलज ऊँचा है।\frac{dy}{dx} bhi hai, par second derivative ooncha hai.
   \text{(ii) highest derivative } \dfrac{d^2y}{dx^2} \Rightarrow \text{order } 2
4. It is a polynomial in y'' and y'; the square is on y', not on the highest derivative.यह y'' और y' में बहुपद है; वर्ग y' पर है, उच्चतम अवकलज पर नहीं।Ye y'' aur y' mein polynomial hai; square y' par hai, highest derivative par nahi.
   \text{(ii) power of } \dfrac{d^2y}{dx^2} \text{ is } 1 \Rightarrow \text{degree } 1
5. The third derivative is the highest present.तृतीय अवकलज सबसे ऊँचा है।Third derivative sabse ooncha hai.
   \text{(iii) highest derivative } y''' \Rightarrow \text{order } 3
6. y' is in an exponent, so the degree is not defined.y' घातांक में है, इसलिए घात परिभाषित नहीं है।y' exponent mein hai, isliye degree defined nahi hai.
   e^{y'} \Rightarrow \text{not a polynomial in derivatives}

✔ (i) order 1, degree 1; (ii) order 2, degree 1; (iii) order 3, degree not defined.(i) कोटि 1, घात 1; (ii) कोटि 2, घात 1; (iii) कोटि 3, घात परिभाषित नहीं।(i) order 1, degree 1; (ii) order 2, degree 1; (iii) order 3, degree defined nahi.

1. Differentiate y = e^{-3x} (chain rule).y = e^{-3x} का अवकलन कीजिए (श्रृंखला नियम)।y = e^{-3x} ko differentiate kijiye (chain rule).
   \dfrac{dy}{dx} = -3e^{-3x}
2. Differentiate once more: (-3)(-3) = 9.एक बार और अवकलन: (-3)(-3) = 9।Ek baar aur differentiate: (-3)(-3) = 9.
   \dfrac{d^2y}{dx^2} = 9e^{-3x}
3. Substitute y'', y' and y in the equation.y'', y' और y को समीकरण में रखिए।y'', y' aur y ko equation mein rakhiye.
   \text{L.H.S.} = 9e^{-3x} + (-3e^{-3x}) - 6e^{-3x}
4. The terms cancel.पद कट जाते हैं।Terms cancel ho jaate hain.
   = 9e^{-3x} - 9e^{-3x} = 0 = \text{R.H.S.}

✔ y = e^{-3x} is a solution of the given differential equation.y = e^{-3x} दिए गए अवकल समीकरण का हल है।y = e^{-3x} diye gaye differential equation ka solution hai.

1. Differentiate term by term.पद-दर-पद अवकलन कीजिए।Term-by-term differentiate kijiye.
   \dfrac{dy}{dx} = -a\sin x + b\cos x
2. Differentiate again.फिर से अवकलन कीजिए।Phir se differentiate kijiye.
   \dfrac{d^2y}{dx^2} = -a\cos x - b\sin x
3. Substitute y'' and y.y'' और y रखिए।y'' aur y rakhiye.
   \text{L.H.S.} = (-a\cos x - b\sin x) + (a\cos x + b\sin x) = 0 = \text{R.H.S.}

✔ y = a\cos x + b\sin x is a solution (for every a, b) — a general solution with two arbitrary constants for this second order equation.y = a\cos x + b\sin x हल है (हर a, b के लिए) — इस द्वितीय कोटि के समीकरण का दो स्वेच्छ अचरों वाला व्यापक हल।y = a\cos x + b\sin x solution hai (har a, b ke liye) — is second order equation ka do arbitrary constants wala general solution.

1. Separate: multiply by (2 - y)\,dx.चर अलग कीजिए: (2 - y)\,dx से गुणा कीजिए।Variables alag kijiye: (2 - y)\,dx se multiply kijiye.
   (2 - y)\,dy = (x + 1)\,dx
2. Integrate both sides.दोनों पक्षों का समाकलन कीजिए।Dono sides integrate kijiye.
   \displaystyle\int (2 - y)\,dy = \int (x + 1)\,dx
3. Power rule on each side; one constant C_1.हर पक्ष पर घात नियम; एक अचर C_1।Har side par power rule; ek constant C_1.
   2y - \dfrac{y^2}{2} = \dfrac{x^2}{2} + x + C_1
4. Multiply by 2 and bring all terms to one side.2 से गुणा कीजिए और सभी पद एक ओर लाइए।2 se multiply kijiye aur saare terms ek taraf laaiye.
   x^2 + y^2 + 2x - 4y + 2C_1 = 0
5. Rename the constant.अचर का नया नाम रखिए।Constant ka naya naam rakhiye.
   x^2 + y^2 + 2x - 4y + C = 0,\quad C = 2C_1

✔ x^2 + y^2 + 2x - 4y + C = 0 — the general solution (a family of circles).x^2 + y^2 + 2x - 4y + C = 0 — व्यापक हल (वृत्तों का कुल)।x^2 + y^2 + 2x - 4y + C = 0 — general solution (circles ki family).

1. Since 1 + y^2 \ne 0, we can divide by it and separate.चूँकि 1 + y^2 \ne 0, इससे भाग देकर चर अलग कर सकते हैं।Kyunki 1 + y^2 \ne 0, isse divide karke variables alag kar sakte hain.
   \dfrac{dy}{1 + y^2} = \dfrac{dx}{1 + x^2}
2. Integrate both sides.दोनों पक्षों का समाकलन कीजिए।Dono sides integrate kijiye.
   \displaystyle\int \dfrac{dy}{1 + y^2} = \int \dfrac{dx}{1 + x^2}
3. Standard integral \int \frac{dt}{1 + t^2} = \tan^{-1} t.मानक समाकल \int \frac{dt}{1 + t^2} = \tan^{-1} t।Standard integral \int \frac{dt}{1 + t^2} = \tan^{-1} t.
   \tan^{-1} y = \tan^{-1} x + C

✔ \tan^{-1} y = \tan^{-1} x + C\tan^{-1} y = \tan^{-1} x + C\tan^{-1} y = \tan^{-1} x + C

1. For y \ne 0, divide by y^2 and separate.y \ne 0 के लिए y^2 से भाग देकर चर अलग कीजिए।y \ne 0 ke liye y^2 se divide karke variables alag kijiye.
   \dfrac{dy}{y^2} = -4x\,dx
2. \int y^{-2}dy = -y^{-1} and \int x\,dx = \frac{x^2}{2}.\int y^{-2}dy = -y^{-1} और \int x\,dx = \frac{x^2}{2}।\int y^{-2}dy = -y^{-1} aur \int x\,dx = \frac{x^2}{2}.
   \displaystyle\int y^{-2}\,dy = -4\int x\,dx \;\Rightarrow\; -\dfrac{1}{y} = -2x^2 + C
3. Take reciprocals of both sides and multiply by -1.दोनों पक्षों का व्युत्क्रम लेकर -1 से गुणा कीजिए।Dono sides ka reciprocal lekar -1 se multiply kijiye.
   y = \dfrac{1}{2x^2 - C} … (2)
4. Put x = 0, y = 1 in (2).(2) में x = 0, y = 1 रखिए।(2) mein x = 0, y = 1 rakhiye.
   1 = \dfrac{1}{0 - C} \Rightarrow C = -1
5. Substitute C = -1 in (2).(2) में C = -1 रखिए।(2) mein C = -1 rakhiye.
   y = \dfrac{1}{2x^2 + 1}

✔ y = \dfrac{1}{2x^2 + 1} — the particular solution.y = \dfrac{1}{2x^2 + 1} — विशिष्ट हल।y = \dfrac{1}{2x^2 + 1} — particular solution.

1. Divide by x (x \ne 0). Here dy and dx are handled like separate quantities (Leibnitz notation).x से भाग दीजिए (x \ne 0)। यहाँ dy और dx को अलग-अलग राशियों की तरह लिया गया है (लाइबनिज़ संकेतन)।x se divide kijiye (x \ne 0). Yahan dy aur dx ko alag quantities ki tarah liya hai (Leibnitz notation).
   dy = \left(2x + \dfrac{1}{x}\right)dx
2. Integrate: \int 2x\,dx = x^2, \int \frac{dx}{x} = \log|x|. This is the family of solution curves.समाकलन: \int 2x\,dx = x^2, \int \frac{dx}{x} = \log|x|। यह हल वक्रों का कुल है।Integrate: \int 2x\,dx = x^2, \int \frac{dx}{x} = \log|x|. Ye solution curves ki family hai.
   y = x^2 + \log|x| + C … (2)
3. The curve passes through (1, 1); \log 1 = 0.वक्र (1, 1) से जाता है; \log 1 = 0।Curve (1, 1) se jaata hai; \log 1 = 0.
   1 = 1 + \log 1 + C \Rightarrow C = 0

✔ y = x^2 + \log|x|y = x^2 + \log|x|y = x^2 + \log|x|

1. The slope of the tangent is \frac{dy}{dx}.स्पर्श रेखा की प्रवणता \frac{dy}{dx} होती है।Tangent ki slope \frac{dy}{dx} hoti hai.
   \dfrac{dy}{dx} = \dfrac{2x}{y^2}
2. Separate the variables.चर अलग कीजिए।Variables alag kijiye.
   y^2\,dy = 2x\,dx
3. Integrate both sides.दोनों पक्षों का समाकलन कीजिए।Dono sides integrate kijiye.
   \dfrac{y^3}{3} = x^2 + C … (3)
4. Put x = -2, y = 3: 9 = 4 + C.x = -2, y = 3 रखिए: 9 = 4 + C।x = -2, y = 3 rakhiye: 9 = 4 + C.
   \dfrac{27}{3} = 4 + C \Rightarrow C = 5
5. Substitute C and solve for y.C रखिए और y ज्ञात कीजिए।C rakhiye aur y nikaaliye.
   \dfrac{y^3}{3} = x^2 + 5 \;\Rightarrow\; y = (3x^2 + 15)^{\frac{1}{3}}

✔ \dfrac{y^3}{3} = x^2 + 5, i.e. y = (3x^2 + 15)^{1/3}\dfrac{y^3}{3} = x^2 + 5, अर्थात् y = (3x^2 + 15)^{1/3}\dfrac{y^3}{3} = x^2 + 5, yaani y = (3x^2 + 15)^{1/3}

1. P = principal at time t; rate of growth is 5% of the principal present.t समय पर मूलधन P; वृद्धि की दर उपस्थित मूलधन का 5% है।t time par principal P; growth ki rate maujood principal ka 5% hai.
   \dfrac{dP}{dt} = \left(\dfrac{5}{100}\right)P = \dfrac{P}{20}
2. Separate the variables.चर अलग कीजिए।Variables alag kijiye.
   \dfrac{dP}{P} = \dfrac{dt}{20}
3. Integrate, then take exponentials (P > 0).समाकलन कीजिए, फिर घातांकी रूप लीजिए (P > 0)।Integrate kijiye, phir exponential lijiye (P > 0).
   \log P = \dfrac{t}{20} + C_1 \;\Rightarrow\; P = e^{\frac{t}{20}}\cdot e^{C_1}
4. Rename the constant.अचर का नया नाम।Constant ka naya naam.
   P = C\,e^{\frac{t}{20}},\quad C = e^{C_1} … (3)
5. P = 1000 when t = 0.t = 0 पर P = 1000।t = 0 par P = 1000.
   1000 = C\,e^{0} \Rightarrow C = 1000
6. Let t years be the doubling time; take \log_e.मान लीजिए t वर्ष में दुगुना; \log_e लीजिए।Maan lijiye t saal mein double; \log_e lijiye.
   2000 = 1000\,e^{\frac{t}{20}} \Rightarrow e^{\frac{t}{20}} = 2 \Rightarrow t = 20\log_e 2

✔ t = 20\log_e 2 years (≈ 13.86 years, using \log_e 2 \approx 0.6931).t = 20\log_e 2 वर्ष (≈ 13.86 वर्ष, \log_e 2 \approx 0.6931 लेकर)।t = 20\log_e 2 saal (≈ 13.86 saal, \log_e 2 \approx 0.6931 lekar).

1. Write it as \frac{dy}{dx} = F(x, y).इसे \frac{dy}{dx} = F(x, y) के रूप में लिखिए।Ise \frac{dy}{dx} = F(x, y) form mein likhiye.
   \dfrac{dy}{dx} = \dfrac{x + 2y}{x - y} = F(x, y) … (1)
2. \lambda cancels completely, so F is homogeneous of degree zero: the equation is homogeneous.\lambda पूरी तरह कट जाता है, इसलिए F शून्य घात का समघातीय फलन है: समीकरण समघातीय है।\lambda poori tarah cancel ho jaata hai, isliye F degree zero ka homogeneous function hai: equation homogeneous hai.
   F(\lambda x, \lambda y) = \dfrac{\lambda(x + 2y)}{\lambda(x - y)} = \lambda^0 F(x, y)
3. Substitute y = vx and divide numerator and denominator by x.y = vx रखिए और अंश-हर को x से भाग दीजिए।y = vx rakhiye aur numerator-denominator ko x se divide kijiye.
   y = vx,\ \ \dfrac{dy}{dx} = v + x\dfrac{dv}{dx} \;\Rightarrow\; v + x\dfrac{dv}{dx} = \dfrac{1 + 2v}{1 - v}
4. Subtract v and simplify: 1 + 2v - v + v^2.v घटाकर सरल कीजिए: 1 + 2v - v + v^2।v ghataakar simplify kijiye: 1 + 2v - v + v^2.
   x\dfrac{dv}{dx} = \dfrac{1 + 2v}{1 - v} - v = \dfrac{v^2 + v + 1}{1 - v}
5. Separate the variables (note the sign: 1 - v = -(v - 1)).चर अलग कीजिए (चिह्न देखिए: 1 - v = -(v - 1))।Variables alag kijiye (sign dekhiye: 1 - v = -(v - 1)).
   \dfrac{v - 1}{v^2 + v + 1}\,dv = -\dfrac{dx}{x}
6. Write v - 1 = \frac12(2v + 1) - \frac32 and complete the square in the denominator.v - 1 = \frac12(2v + 1) - \frac32 लिखिए और हर में पूर्ण वर्ग बनाइए।v - 1 = \frac12(2v + 1) - \frac32 likhiye aur denominator mein perfect square banaiye.
   \dfrac{1}{2}\displaystyle\int \dfrac{2v + 1}{v^2 + v + 1}\,dv - \dfrac{3}{2}\int \dfrac{dv}{\left(v + \frac12\right)^2 + \left(\frac{\sqrt3}{2}\right)^2} = -\log|x| + C_1
7. First integral is a log; the second gives \frac32 \cdot \frac{2}{\sqrt3}\tan^{-1}\frac{2v + 1}{\sqrt3} = \sqrt3\tan^{-1}\frac{2v + 1}{\sqrt3}.पहला समाकल log है; दूसरा \frac32 \cdot \frac{2}{\sqrt3}\tan^{-1}\frac{2v + 1}{\sqrt3} = \sqrt3\tan^{-1}\frac{2v + 1}{\sqrt3} देता है।Pehla integral log hai; doosra \frac32 \cdot \frac{2}{\sqrt3}\tan^{-1}\frac{2v + 1}{\sqrt3} = \sqrt3\tan^{-1}\frac{2v + 1}{\sqrt3} deta hai.
   \dfrac{1}{2}\log(v^2 + v + 1) - \sqrt{3}\tan^{-1}\dfrac{2v + 1}{\sqrt3} = -\log|x| + C_1
8. Move \log|x| = \frac12\log x^2 to the left and combine logs.\log|x| = \frac12\log x^2 को बाईं ओर ले जाकर log मिलाइए।\log|x| = \frac12\log x^2 ko left mein le jaakar logs jodiye.
   \dfrac{1}{2}\log\left[(v^2 + v + 1)x^2\right] = \sqrt3\tan^{-1}\dfrac{2v + 1}{\sqrt3} + C_1
9. Replace v by \frac{y}{x}: (v^2 + v + 1)x^2 = y^2 + xy + x^2.v के स्थान पर \frac{y}{x} रखिए: (v^2 + v + 1)x^2 = y^2 + xy + x^2।v ki jagah \frac{y}{x} rakhiye: (v^2 + v + 1)x^2 = y^2 + xy + x^2.
   \dfrac{1}{2}\log(y^2 + xy + x^2) = \sqrt3\tan^{-1}\left(\dfrac{2y + x}{\sqrt3\,x}\right) + C_1
10. Multiply by 2; C = 2C_1.2 से गुणा कीजिए; C = 2C_1।2 se multiply kijiye; C = 2C_1.
    \log(x^2 + xy + y^2) = 2\sqrt3\tan^{-1}\left(\dfrac{x + 2y}{\sqrt3\,x}\right) + C

✔ \log(x^2 + xy + y^2) = 2\sqrt3\tan^{-1}\left(\dfrac{x + 2y}{\sqrt3\,x}\right) + C\log(x^2 + xy + y^2) = 2\sqrt3\tan^{-1}\left(\dfrac{x + 2y}{\sqrt3\,x}\right) + C\log(x^2 + xy + y^2) = 2\sqrt3\tan^{-1}\left(\dfrac{x + 2y}{\sqrt3\,x}\right) + C

1. Divide by x\cos\left(\frac{y}{x}\right).x\cos\left(\frac{y}{x}\right) से भाग दीजिए।x\cos\left(\frac{y}{x}\right) se divide kijiye.
   \dfrac{dy}{dx} = \dfrac{y\cos\left(\frac{y}{x}\right) + x}{x\cos\left(\frac{y}{x}\right)} = F(x, y)
2. \frac{\lambda y}{\lambda x} = \frac{y}{x}, so the cos part does not change: homogeneous of degree zero.\frac{\lambda y}{\lambda x} = \frac{y}{x}, इसलिए cos वाला भाग नहीं बदलता: शून्य घात का समघातीय।\frac{\lambda y}{\lambda x} = \frac{y}{x}, isliye cos wala part nahi badalta: degree zero ka homogeneous.
   F(\lambda x, \lambda y) = \dfrac{\lambda\left[y\cos\left(\frac{y}{x}\right) + x\right]}{\lambda\left[x\cos\left(\frac{y}{x}\right)\right]} = \lambda^0 F(x, y)
3. Put y = vx, \frac{dy}{dx} = v + x\frac{dv}{dx}.y = vx, \frac{dy}{dx} = v + x\frac{dv}{dx} रखिए।y = vx, \frac{dy}{dx} = v + x\frac{dv}{dx} rakhiye.
   v + x\dfrac{dv}{dx} = \dfrac{v\cos v + 1}{\cos v}
4. Subtract v, then separate.v घटाइए, फिर चर अलग कीजिए।v ghataiye, phir variables alag kijiye.
   x\dfrac{dv}{dx} = \dfrac{1}{\cos v} \;\Rightarrow\; \cos v\,dv = \dfrac{dx}{x}
5. Integrate; writing the constant as \log|C| lets the logs combine.समाकलन कीजिए; अचर को \log|C| लिखने से log मिल जाते हैं।Integrate kijiye; constant ko \log|C| likhne se logs jud jaate hain.
   \sin v = \log|x| + \log|C| = \log|Cx|

✔ \sin\left(\dfrac{y}{x}\right) = \log|Cx|\sin\left(\dfrac{y}{x}\right) = \log|Cx|\sin\left(\dfrac{y}{x}\right) = \log|Cx|

1. Here it is easier to write \frac{dx}{dy}.यहाँ \frac{dx}{dy} लिखना आसान है।Yahan \frac{dx}{dy} likhna aasaan hai.
   \dfrac{dx}{dy} = \dfrac{2x\,e^{\frac{x}{y}} - y}{2y\,e^{\frac{x}{y}}} = F(x, y) … (1)
2. Homogeneous of degree zero, so the equation is homogeneous.शून्य घात का समघातीय, अतः समीकरण समघातीय है।Degree zero ka homogeneous, isliye equation homogeneous hai.
   F(\lambda x, \lambda y) = \dfrac{\lambda\left(2x\,e^{\frac{x}{y}} - y\right)}{\lambda\left(2y\,e^{\frac{x}{y}}\right)} = \lambda^0 F(x, y)
3. Since the equation is in \frac{dx}{dy} form, substitute x = vy.समीकरण \frac{dx}{dy} रूप में है, इसलिए x = vy रखिए।Equation \frac{dx}{dy} form mein hai, isliye x = vy rakhiye.
   x = vy,\ \dfrac{dx}{dy} = v + y\dfrac{dv}{dy} \;\Rightarrow\; v + y\dfrac{dv}{dy} = \dfrac{2v\,e^v - 1}{2e^v}
4. Subtract v and separate.v घटाइए और चर अलग कीजिए।v ghataiye aur variables alag kijiye.
   y\dfrac{dv}{dy} = -\dfrac{1}{2e^v} \;\Rightarrow\; 2e^v\,dv = -\dfrac{dy}{y}
5. Integrate and replace v by \frac{x}{y}.समाकलन कीजिए और v के स्थान पर \frac{x}{y} रखिए।Integrate kijiye aur v ki jagah \frac{x}{y} rakhiye.
   2e^v = -\log|y| + C \;\Rightarrow\; 2e^{\frac{x}{y}} + \log|y| = C … (3)
6. Put x = 0, y = 1.x = 0, y = 1 रखिए।x = 0, y = 1 rakhiye.
   2e^0 + \log 1 = C \Rightarrow C = 2

✔ 2e^{\frac{x}{y}} + \log|y| = 22e^{\frac{x}{y}} + \log|y| = 22e^{\frac{x}{y}} + \log|y| = 2

1. Slope of tangent = \frac{dy}{dx}; divide numerator and denominator by x^2: a function of \frac{y}{x}, so homogeneous.स्पर्श रेखा की प्रवणता = \frac{dy}{dx}; अंश-हर को x^2 से भाग दीजिए: \frac{y}{x} का फलन, अतः समघातीय।Tangent ki slope = \frac{dy}{dx}; numerator-denominator ko x^2 se divide kijiye: \frac{y}{x} ka function, isliye homogeneous.
   \dfrac{dy}{dx} = \dfrac{x^2 + y^2}{2xy} = \dfrac{1 + \frac{y^2}{x^2}}{2\frac{y}{x}}
2. Put y = vx and subtract v.y = vx रखिए और v घटाइए।y = vx rakhiye aur v ghataiye.
   v + x\dfrac{dv}{dx} = \dfrac{1 + v^2}{2v} \;\Rightarrow\; x\dfrac{dv}{dx} = \dfrac{1 - v^2}{2v}
3. Separate; \frac{2v}{1 - v^2} = -\frac{2v}{v^2 - 1}.चर अलग कीजिए; \frac{2v}{1 - v^2} = -\frac{2v}{v^2 - 1}।Variables alag kijiye; \frac{2v}{1 - v^2} = -\frac{2v}{v^2 - 1}.
   \dfrac{2v}{v^2 - 1}\,dv = -\dfrac{dx}{x}
4. Numerator is the derivative of the denominator, so the integral is a log.अंश, हर का अवकलज है, इसलिए समाकल log है।Numerator, denominator ka derivative hai, isliye integral log hai.
   \log|v^2 - 1| = -\log|x| + \log|C_1| \;\Rightarrow\; (v^2 - 1)x = \pm C_1
5. Replace v by \frac{y}{x} and multiply by x.v के स्थान पर \frac{y}{x} रखकर x से गुणा कीजिए।v ki jagah \frac{y}{x} rakhkar x se multiply kijiye.
   \left(\dfrac{y^2}{x^2} - 1\right)x = \pm C_1 \;\Rightarrow\; y^2 - x^2 = \pm C_1 x

✔ x^2 - y^2 = Cx (with C = \mp C_1) — the required family.x^2 - y^2 = Cx (C = \mp C_1) — अभीष्ट कुल।x^2 - y^2 = Cx (C = \mp C_1) — required family.

1. It is of the form \frac{dy}{dx} + Py = Q. Careful: P carries the minus sign.यह \frac{dy}{dx} + Py = Q रूप का है। ध्यान दीजिए: ऋण चिह्न P में है।Ye \frac{dy}{dx} + Py = Q form ka hai. Dhyaan dijiye: minus sign P mein hai.
   P = -1,\quad Q = \cos x
2. Integrating factor.समाकलन गुणक।Integrating factor.
   \text{I.F.} = e^{\int -1\,dx} = e^{-x}
3. Multiply by the I.F.; the L.H.S. becomes an exact derivative. Integrate.I.F. से गुणा कीजिए; बायाँ पक्ष एक अवकलज बन जाता है। समाकलन कीजिए।I.F. se multiply kijiye; L.H.S. ek derivative ban jaati hai. Integrate kijiye.
   \dfrac{d}{dx}\left(y\,e^{-x}\right) = e^{-x}\cos x \;\Rightarrow\; y\,e^{-x} = \displaystyle\int e^{-x}\cos x\,dx + C … (1)
4. By parts with \cos x first and e^{-x} second (\int e^{-x}dx = -e^{-x}).खंडशः समाकलन: पहला \cos x, दूसरा e^{-x} (\int e^{-x}dx = -e^{-x})।By parts: pehla \cos x, doosra e^{-x} (\int e^{-x}dx = -e^{-x}).
   I = \displaystyle\int e^{-x}\cos x\,dx = -e^{-x}\cos x - \int \sin x\,e^{-x}\,dx
5. By parts once more on \int \sin x\,e^{-x}\,dx; the same integral I comes back.\int \sin x\,e^{-x}\,dx पर फिर से खंडशः; वही समाकल I वापस आता है।\int \sin x\,e^{-x}\,dx par phir se by parts; wahi integral I wapas aata hai.
   I = -e^{-x}\cos x + e^{-x}\sin x - I
6. Bring I to the left side.I को बाईं ओर लाइए।I ko left side laaiye.
   2I = (\sin x - \cos x)\,e^{-x} \;\Rightarrow\; I = \dfrac{(\sin x - \cos x)\,e^{-x}}{2}
7. Put I in (1).I को (1) में रखिए।I ko (1) mein rakhiye.
   y\,e^{-x} = \left(\dfrac{\sin x - \cos x}{2}\right)e^{-x} + C

✔ y = \dfrac{\sin x - \cos x}{2} + C\,e^{x}y = \dfrac{\sin x - \cos x}{2} + C\,e^{x}y = \dfrac{\sin x - \cos x}{2} + C\,e^{x}

1. Divide by x so that the coefficient of \frac{dy}{dx} is 1.x से भाग दीजिए ताकि \frac{dy}{dx} का गुणांक 1 हो।x se divide kijiye taaki \frac{dy}{dx} ka coefficient 1 ho.
   \dfrac{dy}{dx} + \dfrac{2}{x}\,y = x
2. Read off P and Q.P और Q पहचानिए।P aur Q pehchaaniye.
   P = \dfrac{2}{x},\quad Q = x
3. Use e^{\log f(x)} = f(x).e^{\log f(x)} = f(x) का प्रयोग कीजिए।e^{\log f(x)} = f(x) use kijiye.
   \text{I.F.} = e^{\int \frac{2}{x}dx} = e^{2\log x} = e^{\log x^2} = x^2
4. y\,(\text{I.F.}) = \int Q\,(\text{I.F.})\,dx + C.y\,(\text{I.F.}) = \int Q\,(\text{I.F.})\,dx + C।y\,(\text{I.F.}) = \int Q\,(\text{I.F.})\,dx + C.
   y\cdot x^2 = \displaystyle\int (x)(x^2)\,dx + C = \int x^3\,dx + C = \dfrac{x^4}{4} + C

✔ y = \dfrac{x^2}{4} + C\,x^{-2}y = \dfrac{x^2}{4} + C\,x^{-2}y = \dfrac{x^2}{4} + C\,x^{-2}

1. Divide by y\,dy: it is linear in x (the form \frac{dx}{dy} + P_1 x = Q_1).y\,dy से भाग दीजिए: यह x में रैखिक है (\frac{dx}{dy} + P_1 x = Q_1 रूप)।y\,dy se divide kijiye: ye x mein linear hai (\frac{dx}{dy} + P_1 x = Q_1 form).
   \dfrac{dx}{dy} - \dfrac{x}{y} = 2y
2. Read off P_1 and Q_1.P_1 और Q_1 पहचानिए।P_1 aur Q_1 pehchaaniye.
   P_1 = -\dfrac{1}{y},\quad Q_1 = 2y
3. Integrate with respect to y.y के सापेक्ष समाकलन।y ke respect mein integrate.
   \text{I.F.} = e^{\int -\frac{1}{y}dy} = e^{-\log y} = e^{\log (y)^{-1}} = \dfrac{1}{y}
4. x\,(\text{I.F.}) = \int Q_1\,(\text{I.F.})\,dy + C.x\,(\text{I.F.}) = \int Q_1\,(\text{I.F.})\,dy + C।x\,(\text{I.F.}) = \int Q_1\,(\text{I.F.})\,dy + C.
   x\cdot\dfrac{1}{y} = \displaystyle\int (2y)\left(\dfrac{1}{y}\right)dy + C = \int 2\,dy + C = 2y + C

✔ x = 2y^2 + Cyx = 2y^2 + Cyx = 2y^2 + Cy

1. Already in the form \frac{dy}{dx} + Py = Q.पहले से \frac{dy}{dx} + Py = Q रूप में है।Pehle se \frac{dy}{dx} + Py = Q form mein hai.
   P = \cot x,\quad Q = 2x + x^2\cot x
2. \int \cot x\,dx = \log \sin x.\int \cot x\,dx = \log \sin x।\int \cot x\,dx = \log \sin x.
   \text{I.F.} = e^{\int \cot x\,dx} = e^{\log \sin x} = \sin x
3. (2x + x^2\cot x)\sin x = 2x\sin x + x^2\cos x.(2x + x^2\cot x)\sin x = 2x\sin x + x^2\cos x।(2x + x^2\cot x)\sin x = 2x\sin x + x^2\cos x.
   y\sin x = \displaystyle\int 2x\sin x\,dx + \int x^2\cos x\,dx + C
4. Integrate the first integral by parts with \sin x as the first function: \int 2x\sin x\,dx = x^2\sin x - \int x^2\cos x\,dx.पहले समाकल का खंडशः समाकलन, \sin x को पहला फलन लेकर: \int 2x\sin x\,dx = x^2\sin x - \int x^2\cos x\,dx।Pehle integral ko by parts, \sin x ko first function lekar: \int 2x\sin x\,dx = x^2\sin x - \int x^2\cos x\,dx.
   y\sin x = x^2\sin x - \displaystyle\int x^2\cos x\,dx + \int x^2\cos x\,dx + C
5. The two integrals cancel.दोनों समाकल कट जाते हैं।Dono integrals cancel ho jaate hain.
   y\sin x = x^2\sin x + C … (1)
6. Put y = 0, x = \frac{\pi}{2}.y = 0, x = \frac{\pi}{2} रखिए।y = 0, x = \frac{\pi}{2} rakhiye.
   0 = \left(\dfrac{\pi}{2}\right)^2\sin\dfrac{\pi}{2} + C \;\Rightarrow\; C = -\dfrac{\pi^2}{4}
7. Substitute C in (1).(1) में C रखिए।(1) mein C rakhiye.
   y\sin x = x^2\sin x - \dfrac{\pi^2}{4}

✔ y = x^2 - \dfrac{\pi^2}{4\sin x} (\sin x \ne 0)y = x^2 - \dfrac{\pi^2}{4\sin x} (\sin x \ne 0)y = x^2 - \dfrac{\pi^2}{4\sin x} (\sin x \ne 0)

1. Slope = abscissa + abscissa × ordinate.प्रवणता = भुज + भुज × कोटि।Slope = abscissa + abscissa × ordinate.
   \dfrac{dy}{dx} = x + xy \;\Rightarrow\; \dfrac{dy}{dx} - xy = x
2. Linear with P = −x.P = −x के साथ रैखिक।P = −x ke saath linear.
   P = -x,\quad Q = x,\quad \text{I.F.} = e^{\int -x\,dx} = e^{-\frac{x^2}{2}}
3. y\,(\text{I.F.}) = \int Q\,(\text{I.F.})\,dx + C.y\,(\text{I.F.}) = \int Q\,(\text{I.F.})\,dx + C।y\,(\text{I.F.}) = \int Q\,(\text{I.F.})\,dx + C.
   y\,e^{-\frac{x^2}{2}} = \displaystyle\int x\,e^{-\frac{x^2}{2}}\,dx + C … (2)
4. Substitution.प्रतिस्थापन।Substitution.
   -\dfrac{x^2}{2} = t,\ x\,dx = -dt \;\Rightarrow\; \displaystyle\int x\,e^{-\frac{x^2}{2}}\,dx = -\int e^t\,dt = -e^{-\frac{x^2}{2}}
5. Multiply by e^{\frac{x^2}{2}}. This is the family of curves.e^{\frac{x^2}{2}} से गुणा कीजिए। यह वक्रों का कुल है।e^{\frac{x^2}{2}} se multiply kijiye. Ye curves ki family hai.
   y\,e^{-\frac{x^2}{2}} = -e^{-\frac{x^2}{2}} + C \;\Rightarrow\; y = -1 + C\,e^{\frac{x^2}{2}} … (3)
6. The curve passes through (0, 1).वक्र (0, 1) से जाता है।Curve (0, 1) se jaata hai.
   1 = -1 + C\,e^0 \Rightarrow C = 2

✔ y = -1 + 2\,e^{\frac{x^2}{2}} (the orange curve in the slope-field picture above).y = -1 + 2\,e^{\frac{x^2}{2}} (ऊपर ढाल-क्षेत्र वाले चित्र का नारंगी वक्र)।y = -1 + 2\,e^{\frac{x^2}{2}} (upar slope-field wali picture ka orange curve).

1. Take e^{ax} common.e^{ax} उभयनिष्ठ लीजिए।e^{ax} common lijiye.
   y = e^{ax}\left[c_1\cos bx + c_2\sin bx\right] … (1)
2. Product rule: e^{ax}[-bc_1\sin bx + bc_2\cos bx] + a e^{ax}[c_1\cos bx + c_2\sin bx], then collect cos and sin.गुणनफल नियम: e^{ax}[-bc_1\sin bx + bc_2\cos bx] + a e^{ax}[c_1\cos bx + c_2\sin bx], फिर cos और sin के पद एकत्र कीजिए।Product rule: e^{ax}[-bc_1\sin bx + bc_2\cos bx] + a e^{ax}[c_1\cos bx + c_2\sin bx], phir cos aur sin ke terms ikattha kijiye.
   \dfrac{dy}{dx} = e^{ax}\left[(bc_2 + ac_1)\cos bx + (ac_2 - bc_1)\sin bx\right] … (2)
3. Differentiate (2) the same way and collect terms.(2) का इसी तरह अवकलन करके पद एकत्र कीजिए।(2) ko isi tarah differentiate karke terms ikattha kijiye.
   \dfrac{d^2y}{dx^2} = e^{ax}\left[(a^2c_2 - 2abc_1 - b^2c_2)\sin bx + (a^2c_1 + 2abc_2 - b^2c_1)\cos bx\right]
4. Coefficient of \sin bx in L.H.S. after substituting.रखने के बाद बाएँ पक्ष में \sin bx का गुणांक।Rakhne ke baad L.H.S. mein \sin bx ka coefficient.
   \sin bx:\ (a^2c_2 - 2abc_1 - b^2c_2) - 2a(ac_2 - bc_1) + (a^2 + b^2)c_2 = 0
5. Coefficient of \cos bx; everything cancels.\cos bx का गुणांक; सब कट जाता है।\cos bx ka coefficient; sab cancel ho jaata hai.
   \cos bx:\ (a^2c_1 + 2abc_2 - b^2c_1) - 2a(bc_2 + ac_1) + (a^2 + b^2)c_1 = 0
6. Hence verified.अतः सत्यापित।Isliye verify ho gaya.
   \text{L.H.S.} = e^{ax}\left[0\cdot\sin bx + 0\cdot\cos bx\right] = 0 = \text{R.H.S.}

✔ The given function is a solution of the given differential equation.दिया गया फलन दिए गए अवकल समीकरण का हल है।Diya gaya function diye gaye differential equation ka solution hai.

1. Take the antilog (exponential) of both sides.दोनों पक्षों का प्रतिलघुगणक (घातांक) लीजिए।Dono sides ka antilog (exponential) lijiye.
   \dfrac{dy}{dx} = e^{3x + 4y} = e^{3x}\cdot e^{4y} … (1)
2. Separate the variables.चर अलग कीजिए।Variables alag kijiye.
   \dfrac{dy}{e^{4y}} = e^{3x}\,dx \;\Rightarrow\; \displaystyle\int e^{-4y}\,dy = \int e^{3x}\,dx
3. Integrate and multiply by −12.समाकलन करके −12 से गुणा कीजिए।Integrate karke −12 se multiply kijiye.
   \dfrac{e^{-4y}}{-4} = \dfrac{e^{3x}}{3} + C \;\Rightarrow\; 4e^{3x} + 3e^{-4y} + 12C = 0 … (2)
4. Put x = 0, y = 0 in (2).(2) में x = 0, y = 0 रखिए।(2) mein x = 0, y = 0 rakhiye.
   4 + 3 + 12C = 0 \Rightarrow C = -\dfrac{7}{12}

✔ 4e^{3x} + 3e^{-4y} - 7 = 04e^{3x} + 3e^{-4y} - 7 = 04e^{3x} + 3e^{-4y} - 7 = 0

1. Collect the dy terms on the left and the dx terms on the right.dy वाले पद बाईं ओर और dx वाले पद दाईं ओर एकत्र कीजिए।dy wale terms left mein aur dx wale terms right mein ikattha kijiye.
   \left[xy\sin\left(\tfrac{y}{x}\right) - x^2\cos\left(\tfrac{y}{x}\right)\right]dy = \left[xy\cos\left(\tfrac{y}{x}\right) + y^2\sin\left(\tfrac{y}{x}\right)\right]dx
2. Divide numerator and denominator by x^2: a function of \frac{y}{x}, so homogeneous.अंश-हर को x^2 से भाग दीजिए: \frac{y}{x} का फलन, अतः समघातीय।Numerator-denominator ko x^2 se divide kijiye: \frac{y}{x} ka function, isliye homogeneous.
   \dfrac{dy}{dx} = \dfrac{\frac{y}{x}\cos\left(\frac{y}{x}\right) + \frac{y^2}{x^2}\sin\left(\frac{y}{x}\right)}{\frac{y}{x}\sin\left(\frac{y}{x}\right) - \cos\left(\frac{y}{x}\right)} … (1)
3. Put y = vx and subtract v.y = vx रखिए और v घटाइए।y = vx rakhiye aur v ghataiye.
   v + x\dfrac{dv}{dx} = \dfrac{v\cos v + v^2\sin v}{v\sin v - \cos v} \;\Rightarrow\; x\dfrac{dv}{dx} = \dfrac{2v\cos v}{v\sin v - \cos v}
4. Separate; split the fraction: \frac{v\sin v}{v\cos v} - \frac{\cos v}{v\cos v} = \tan v - \frac1v.चर अलग कीजिए; भिन्न को तोड़िए: \frac{v\sin v}{v\cos v} - \frac{\cos v}{v\cos v} = \tan v - \frac1v।Variables alag kijiye; fraction todiye: \frac{v\sin v}{v\cos v} - \frac{\cos v}{v\cos v} = \tan v - \frac1v.
   \left(\dfrac{v\sin v - \cos v}{v\cos v}\right)dv = \dfrac{2\,dx}{x} \;\Rightarrow\; \displaystyle\int \tan v\,dv - \int \dfrac{1}{v}\,dv = 2\int \dfrac{1}{x}\,dx
5. \int \tan v\,dv = \log|\sec v|; combine the logs.\int \tan v\,dv = \log|\sec v|; log मिलाइए।\int \tan v\,dv = \log|\sec v|; logs jodiye.
   \log|\sec v| - \log|v| = 2\log|x| + \log|C_1| \;\Rightarrow\; \dfrac{\sec v}{v x^2} = \pm C_1 … (3)
6. Replace v by \frac{y}{x}; C = \pm C_1.v के स्थान पर \frac{y}{x}; C = \pm C_1।v ki jagah \frac{y}{x}; C = \pm C_1.
   \dfrac{\sec\left(\frac{y}{x}\right)}{\left(\frac{y}{x}\right)x^2} = C \;\Rightarrow\; \sec\left(\dfrac{y}{x}\right) = Cxy

✔ \sec\left(\dfrac{y}{x}\right) = Cxy\sec\left(\dfrac{y}{x}\right) = Cxy\sec\left(\dfrac{y}{x}\right) = Cxy

1. Divide by (1 + y^2)\,dy: linear in x.(1 + y^2)\,dy से भाग दीजिए: x में रैखिक।(1 + y^2)\,dy se divide kijiye: x mein linear.
   \dfrac{dx}{dy} + \dfrac{x}{1 + y^2} = \dfrac{\tan^{-1}y}{1 + y^2} … (1)
2. Form \frac{dx}{dy} + P_1 x = Q_1.रूप \frac{dx}{dy} + P_1 x = Q_1।Form \frac{dx}{dy} + P_1 x = Q_1.
   P_1 = \dfrac{1}{1 + y^2},\quad Q_1 = \dfrac{\tan^{-1}y}{1 + y^2},\quad \text{I.F.} = e^{\int \frac{dy}{1 + y^2}} = e^{\tan^{-1}y}
3. x\,(\text{I.F.}) = \int Q_1\,(\text{I.F.})\,dy + C.x\,(\text{I.F.}) = \int Q_1\,(\text{I.F.})\,dy + C।x\,(\text{I.F.}) = \int Q_1\,(\text{I.F.})\,dy + C.
   x\,e^{\tan^{-1}y} = \displaystyle\int \left(\dfrac{\tan^{-1}y}{1 + y^2}\right)e^{\tan^{-1}y}\,dy + C … (2)
4. Substitute, then integrate by parts.प्रतिस्थापन, फिर खंडशः समाकलन।Substitution, phir by parts.
   \tan^{-1}y = t,\ \dfrac{dy}{1 + y^2} = dt:\quad I = \displaystyle\int t\,e^t\,dt = t\,e^t - e^t = e^t(t - 1)
5. Put I = e^{\tan^{-1}y}(\tan^{-1}y - 1) in (2).(2) में I = e^{\tan^{-1}y}(\tan^{-1}y - 1) रखिए।(2) mein I = e^{\tan^{-1}y}(\tan^{-1}y - 1) rakhiye.
   x\,e^{\tan^{-1}y} = e^{\tan^{-1}y}\left(\tan^{-1}y - 1\right) + C

✔ x = (\tan^{-1}y - 1) + C\,e^{-\tan^{-1}y}x = (\tan^{-1}y - 1) + C\,e^{-\tan^{-1}y}x = (\tan^{-1}y - 1) + C\,e^{-\tan^{-1}y}