Questions and Answers #1 Differential equations

Questions and Answers Sheet 1 Differential equations Question #1 Solve the given differential equation by separation of variables. 𝑑𝑆 𝑑𝑟 = 𝑘𝑆 Answer: 𝑑𝑆 = 𝑘𝑆 𝑑𝑟 ⇒ 𝑑𝑆 = 𝑘𝑑𝑟 𝑆 integrating on both the side, ln(𝑆) = 𝑘𝑟 + 𝑐𝑐 is integration constant. ⇒ 𝑆 = 𝐶 × 𝑒 𝑘𝑟 𝐶 is integration constant. Question #2 Find the general solution of the given differential equation. 𝑥 2 𝑦 ′ + 𝑥(𝑥 + 7)𝑦 = 𝑒 𝑥 𝑦= Answer: Consider the differential equation 𝑥 2 𝑦 ′ + 𝑥(𝑥 + 7)𝑦 = 𝑒 𝑥 On rewriting it, we get 2 𝑒𝑥 𝑥 𝑥2 𝑦 ′ + (1 + ) 𝑦 = ...(1) Standard form of the linear differential equation 𝑑𝑦 + 𝑃(𝑥)𝑦 = 𝑓(𝑥) 𝑑𝑥 Clearly equation (1) is in the standard form Compare the differential equation (1) with standard orm, and identity 𝑃(𝑥) = 2+2 𝑥 and 𝑓(𝑥) = The dunctions 𝑃(𝑥) = 𝑒𝑥 𝑥2 2+2 𝑥 The integrating facrote is 2 𝑒 ∫ 𝑃(𝑥)𝑑𝑥 = 𝑒 ∫(1+𝑥)𝑑𝑥 = 𝑒 𝑥+2 ln 𝑥 𝑒 𝑥 𝑒 ln 𝑥 2 and 𝑓(𝑥) = 𝑒𝑥 𝑥2 are continuous on (0, ∞) = 𝑥2𝑒 𝑥 Multiply the standart form, with the integrating factor 2 𝑒𝑥 𝑥 𝑥2 𝑥 2 𝑒 𝑥 (𝑦 ′ + (1 + ) 𝑦) = 𝑥 2 𝑒 𝑥 ⋅ 𝑥 2 𝑒 𝑥 𝑦 ′ + 𝑥 2 𝑒 𝑥 𝑦 + 2𝑥𝑒 𝑥 𝑦 = 𝑒 2𝑥 𝑥 2𝑒 𝑥 𝑑𝑦 𝑑 𝑥 𝑑 2 (𝑒 ) + 𝑒 𝑥 𝑦 (𝑥 ) = 𝑒 2𝑥 + 𝑥 2𝑦 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑 2 𝑥 (𝑥 𝑒 𝑦) = 𝑒 2𝑥 𝑑𝑥 𝑥 2 𝑒 𝑥 𝑦 = ∫ 𝑒 2𝑥 𝑑𝑥 𝑥 2𝑒 𝑥 𝑦 = 1 2𝑥 𝑒 +𝑐 2 𝑥 2𝑒 𝑥 𝑦 = 1 𝑥 𝑒 + 𝑐𝑥 −2 𝑒 −𝑥 2𝑥 2 Hence the general solution of the given differential equation is 𝑦(𝑥) = 1 2𝑥 2 𝑒 𝑥 + 𝑐𝑥 −2 𝑒 −𝑥 ; for 0 < 𝑥 < ∞ That is, general solution is definrd on the interval 𝐼 = (0, ∞) Question #3 Solve the given differential equation by undetermined coefficients 𝑦 ′′ + 7𝑦 ′ + 6𝑦 = 30 Answer: Consider the differential equation 𝑦 ′′ + 7𝑦 ′ + 6𝑦 = 30 ...(1) Rewrite the equation, (𝐷2 + 7𝐷 + 6)𝑦 = 30 where 𝐷 = 𝑑 𝑑𝑥 Auxiliary equation is, 𝑓(𝑚) = 0 𝑚3 + 7𝑚 + 6 = 0 𝑚2 + 6𝑚 + 𝑚 + 6 = 0 𝑚(𝑚 + 6) + 1(𝑚 + 6) = 0 (𝑚 + 6)(𝑚 + 1) = 0 𝑚 = −1, −6 Hence the complementary solution is, 𝑦𝑐 = 𝑐1 𝑒 −𝑥 + 𝑐2 𝑒 −6𝑥 Question #4 Solve the given differential equation by undetermined coefficients 𝑦 ′′ − 14′ + 49𝑦 = 35𝑥 + 4 Answer: Consider the differential equation 𝑦 ′′ − 14′ + 49𝑦 = 35𝑥 + 4 ...(1) Rewrite the equation, (𝐷2 − 14𝐷 + 49)𝑦 = 30 where 𝐷 = 𝑑 𝑑𝑥 Auxiliary equation is, 𝑓(𝑚) = 0 𝑚2 − 14𝑚 + 49 = 0 (𝑚 − 7)2 = 0 𝑚 = 7,7 Hence the complementary solution is, 𝑦𝑐 = 𝑐1 𝑒 7𝑥 + 𝑐2 𝑒 7𝑥 Question #5 𝑥 sin 𝑦 − 𝑦 cos 𝑥 = 𝑐 Answer: Now, differential both sides w.r.y. x, 𝑑 𝑑(𝑐) (𝑥 sin 𝑦 − 𝑦 cos 𝑥) = 𝑑𝑥 𝑑𝑥 𝑑 𝑑 (𝑥 sin 𝑦) − (𝑦 cos 𝑥) = 0 𝑑𝑥 𝑑𝑥 𝑥 𝑑 𝑑(𝑥) 𝑑 𝑑𝑎 sin 𝑦 + sin − (𝑦 cos 𝑥 + cos 𝑥 ) = 0 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑥(cos 𝑦) 𝑑𝑦 𝑑𝑦 + sin 𝑦 − 𝑦(− sin 𝑥) − cos 𝑥 =0 𝑑𝑥 𝑑𝑥 𝑑𝑦 (𝑥 cos 𝑦 − cos 𝑥) = − sin 𝑦 − 𝑦 sin 𝑥 𝑑𝑥 𝑑𝑦(𝑥 cos 𝑦 − cos 𝑥) = −(sin 𝑦 + 𝑦 sin 𝑥)𝑑𝑥 𝑑𝑦(𝑥 cos 𝑦 − cos 𝑥) + (sin 𝑦 + 𝑦 sin 𝑥)𝑑𝑥 = 0 Question #6 Solve the following linear differential equation 𝑦 6 − 64𝑦 = 0 Answer: 𝑦 6 − 64𝑦 = 0 𝑑6 𝑦 − 64𝑥 = 0 𝑑𝑥 6 (𝐷6 − 64)𝑦 = 0 So for C.F. 𝐷6 − 64 = 0 𝐷6 = 64 𝐷 6 = 26 𝐷 So 𝐷 = ±2 So 𝑦 = 𝑐1 𝑒62𝑥 + 𝑐2 𝑒 −2𝑥 Question #7 Integrating factor for 𝑥𝑦𝑑𝑥 + (2𝑥 2 + 3𝑦 2 − 20)𝑑𝑦 = 0 Answer: 𝑥𝑦𝑑𝑥 + (2𝑥 2 + 3𝑦 2 − 20)𝑑𝑦 = 0 𝑚 𝑑𝑥 + 𝑛 𝑑𝑦 = 0 𝜕𝑚 𝜕𝑛 =𝑥 = 4𝑥 𝜕𝑦 𝜕𝑥 Since 𝜕𝑚 𝜕𝑦 ≠ 𝜕𝑛 𝜕𝑥 So not an exact diff. So 𝑚(𝑦) = 𝜕𝑛 𝜕𝑚 − 𝜕𝑥 𝜕𝑦 𝑚 = 4𝑥−𝑥 𝑥𝑦 = 3 𝑦 3 So 𝐼. 𝑓 = 𝑒 ∫ 𝑚(𝑦)𝑑𝑦 = 𝑒 ∫4 𝐼. 𝑓 = 𝑒 3 ln 𝑦 = 𝑒 ln 𝑦 3 𝐼. 𝑓 = 𝑦 3 Question #8 Find the logistic differential equation Answer: Step 1 Given differential equations is 𝑑𝑦/𝑑𝑥 = 𝑘𝑦(1 − 𝑦/𝐿) It can be solved as Step 2 𝑑𝑦 𝑦 = 𝑘𝑦 (1 − ) 𝑑𝑥 𝐿 ⇒ 𝑑𝑦 𝐿−𝑦 = 𝑘𝑦 ( ) 𝑑𝑥 𝐿 ⇒ 𝑑𝑦 𝐾 = 𝑦(𝐿 − 𝑦) 𝑑𝑥 𝐿 ⇒ 𝑑𝑦 𝐾 = 𝑦(𝐿 − 𝑦) 𝑑𝑥 𝐿 ⇒ 𝑑𝑦 𝐾 = − 𝑑𝑥 𝑦(𝑦 − 𝐿) 𝐿 𝑑𝑦 𝑑𝑥 𝑦 = 𝑘𝑦 (1 − ) 𝐿 Integrating both side, we get ⇒∫ 𝑑𝑦 𝐾 = − ∫ 𝑑𝑥 𝑦(𝑦 − 𝐿) 𝐿 1 1 1 𝐾 ⇒ ∫( − ) 𝑑𝑦 = − 𝑥 + 𝑐 𝐿 𝑦−𝐿 𝑦 𝐿  1 y−L K ln | |= − x + c L y L ⇒ ln 𝑦−𝐿 𝑦 = −𝑘𝑥 + 𝑐1 Solution of logistic equation is ln 𝑦−𝐿 𝑦 = −𝑘𝑥 + 𝑐1 Question #9 𝑑2𝑥/𝑑𝑡2 + 4𝑑𝑥/𝑑𝑡 + 4𝑥 = 0 𝑥(0) = 2, 𝑥 ′ (0) = 1 Answer: 𝑑2𝑥 𝑑𝑥 +4 + 4𝑥 = 0 𝑑+2 𝑑𝑡 The equation 𝑟 2 + 4𝑟 + 4 = 0 (𝑟 + 2)2 = 0 𝑟 = −2, −2 The general equation 𝑥 = 𝑐1 𝑒 −2𝑡 + 𝑐2 𝑒 −2𝑡 𝑥(0) = 𝑐1 𝑒 0 2 = 𝑐1 𝑒𝑞(1) 𝑥 1 = −2𝑐1 𝑒 −2𝑡 + 𝑥(−2)𝑐2 𝑒 −2𝑡 + 𝑐2 𝑒 −2𝑡 𝑥 1 (0) = −2𝑐1 + 0 + 𝑐2 1 = −2𝑐1 + 𝑐2 𝑒𝑞(2) Putting the value eqn (1) we get 1 = −2(2) + 𝑐2 1 = −4 + 𝑐2 𝑐2 = 5 Final equation 𝑥 = 2𝑒 −2𝑡 + 5𝑡𝑐2 𝑒 −2𝑡 Question #10 Solve the given LDE: 𝑦 6 − 64𝑦 = 0 Answer: In this question we will use basic of calculus i.e concept of differential equations. We will simply form the auxiliary equation and then we will find the roots of that equation after that we will write the solution of differential equation. Solution of the above question is in the step below. 𝑦 6 − 64𝑦 = 0 (𝐷6 − 64)𝑦 = 0 where 𝐷 = 𝑑 𝑑𝑥 𝑚6 − 64 = 0 (𝑚2 )3 − (4)3 = 0 (𝑚3 )2 − (8)2 = 0 (𝑚3 + 8)(𝑚3 − 8) = 0 (𝑚 + 2)(𝑚2 − 2𝑚 + 4)(𝑚 − 2)(𝑚2 + 2𝑚 + 4) = 0 So, 𝑚 = −2, −2, −2, 2, 2, 2 . So, m has two roots and supeated there times each. Hence, solution of differential equation is 𝑦 = (𝑐1 + 𝑐2 𝑥 + 𝑐3 𝑥 2 )𝑒 −2𝑥 + (𝑐4 + 𝑐5 𝑥 + 𝑐6 𝑥 2 )𝑒 2𝑥 Question #11 2𝑦𝑑𝑥 + 3𝑥𝑑𝑦 = 𝑥𝑦 3 𝑑𝑦; 𝑥 = 1 𝑦 = 1 Answer: Step 1 we have to solve the given differential equations with intial condition. We can solve it variable seperable Step 2 Given 2𝑦𝑑𝑥 + 3𝑥𝑑𝑦 = 𝑥𝑦 3 𝑑𝑦; 𝑥 = 1 𝑦 = 1 ⇒ 2𝑦𝑑𝑥 = 𝑥𝑦 3 𝑑𝑦 − 3𝑥𝑑𝑦 ⇒ 2𝑦𝑑𝑥 = 𝑥(𝑦 3 − 3)𝑑𝑦 ⇒∫ 2𝑑𝑥 𝑦3 − 3 = ∫( ) 𝑑𝑦 𝑥 𝑦 ⇒ 2∫ 𝑑𝑥 3 = ∫ (𝑦 2 − ) 𝑑𝑦 𝑥 𝑦 ⇒ 2 ln(𝑥) = 𝑦3 − 3 ln(𝑦) + 𝑐(1) 3 Put 𝑥 = 1, 𝑦 = 1 ⇒ 2 ln(1) = 13 − 3 ln(1) + 𝑐 3 ⇒ 0=1−0+𝑐 ∴ 𝑐 = −1 Put 𝑐 = −1 in (1) 2 ln(𝑥) = (1) ∫ 𝑑𝑥 𝑥 𝑦3 3 − 3 ln(𝑦) − 1 = 𝑙𝑛(𝑥) + 𝑐 (2) ∫ 𝑥 𝑛 𝑑𝑥 = 𝑥 𝑛+1 𝑛+1 +𝑐 Question #12 State the order of the partial differential equation 𝑢𝑥𝑥 + 𝑢𝑦 = 𝑢 + 4 . Then classify whether it is elliptic, hyperbolic or parabolic. Answer: Step 1 We are given the following partial differential equations 𝑢𝑥𝑥 + 𝑢𝑦 = 𝑢 + 4 . The order of the PDE is 2. Now, we need to classify whether it is elliptic, hyperbolic, or parabolic. Here, 𝐵 = 0, 𝐴 = 1, and 𝐶 = 0 . So, 𝐵2 − 4𝐴𝐶 = 0 . Step 2 Thus, the PDE is parabolic. Question #13 Verify that the function 𝑦 = 𝑒 −3𝑥 is a solution to the differential equation 𝑑 2 𝑦/𝑑𝑥 2 + 𝑑𝑦/𝑑𝑥 − 6𝑦 = 0? Answer: Step 1 To verify the function 𝑦 = 𝑒 −3𝑥 is a solution of the differential equations That is the function 𝑦 = 𝑒 −3𝑥 satisfies the equation 𝑑2𝑦 𝑑𝑥 2 + 𝑑𝑦 𝑑𝑥 𝑑2𝑦 𝑑𝑥 2 = 𝑑𝑦 𝑑𝑥 𝑑 𝑑𝑥 = 𝑑 𝑑𝑥 (𝑒 −3𝑥 ) = 𝑒 −3𝑥 (−3) = −3𝑒 −3𝑥 − 3(𝑒 −3𝑥 ) = (−3) −6𝑦 = −6𝑒 −3𝑥 Therefore, 𝑑2𝑦 𝑑𝑥 2 + 𝑑𝑦 𝑑𝑥 − 6𝑦 = 9𝑒 −3𝑥 − 3𝑒 −3𝑥 − 6𝑒 −3𝑥 = 9𝑒 −3𝑥 − 9𝑒 −3𝑥 =0 𝑑 𝑑𝑥 𝑑𝑥 2 6𝑦 = 0 ...(1) Step 2 Now, 𝑑2 𝑦 (𝑒 −3𝑥 ) = −3(𝑒 −3𝑥 (−3)) = 9𝑒 −3𝑥 + 𝑑𝑦 𝑑𝑥 − 6𝑦 = 0 . Hence, equation (1) satisfies. Therefore, the given function 𝑦 = 𝑒 −3𝑥 is a solution to the given differential equation. Question #14 What is the differential equation of the orthogonal trajectories of the family of curves 𝑥 2 − 3𝑥𝑦 − 2𝑦 2 = 𝐶 ? Answer: Introduction: This is a simple question based on differential equations which could be solved using the basic knowledge of the topic as Solution: Given: 𝑥 2 − 3𝑥𝑦 − 2𝑦 2 = 𝐶 Now differentiating w.r.t. x we get 2𝑥 − 3(𝑦 + 𝑥𝑦 ′ ) − 2(2𝑦)𝑦 ′ = 0 as (𝑥𝑦)′ = 𝑥 ′ 𝑦 + 𝑥𝑦 ′ and (𝐶)′ = 0 2𝑥 − 3𝑦 − 3𝑥𝑦 ′ − 4𝑦𝑦 ′ = 0 or 2𝑥 − 3𝑦 − 𝑦 ′ (3𝑥 + 4𝑦) = 0 Hence 2𝑥 − 3𝑦 − 𝑦 ′ (3𝑥 + 4𝑦) = 0 is the required differential equation. Question #15 (𝐷3 − 14𝐷 + 8)𝑦 = 0 Answer: Step 1. Differential equation Homogeneous DE with constant coefficient- auxiliary equations: (𝐷3 − 14𝐷 + 8)𝑦 = 0 Differential equation: In mathematics, a differential equation is an equation that relates one or more functions and their derivatives. In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two. Such relations are common; therefore, differential equations play a prominent role in many disciplines including engineering, physics, economics, and biology. Step 2. Solution 𝐷=𝑚 𝑚3 − 14𝑚 + 8 = 0 𝑚 = −4 is a factor, ⇒ 𝑚3 + 4𝑚2 − 4𝑚2 − 16𝑚 + 2𝑚 + 8 = 0 ⇒ 𝑚2 (𝑛 + 4) − 4𝑚(𝑚 + 4) + 2(𝑚 + 4) = 0 ⇒ (𝑚 + 4)(𝑚2 − 4𝑚 + 2) = 0 𝑚 = −4 , 𝑚= 4 ± √16 − 8 2 𝑚= 4 ± 2√2 2 𝑚 = 2 ± √2 Step 3. Solution The complementary function is: 𝑦𝑐 = 𝑐1 𝑒 −4𝑥 + 𝑒 2𝑥 (𝑐2 cos √2𝑥 + 𝑐3 sin √2𝑥) 𝑐1 , 𝑐2 , 𝑐3 are arbitrary constant. Question #16 Solve the ff Differential equations. Indicate the type of differential equation 𝑦 ln 𝑥 ln 𝑦 𝑑𝑥 + 𝑑𝑦 = 0 Answer: Step 1 Given: 𝑦(ln 𝑥)(ln 𝑦)𝑑𝑥 + 𝑑𝑦 = 0 ⇒ (ln 𝑥)𝑑𝑥 = −𝑑𝑦/𝑦(ln 𝑦) The above differential equation is of variable separation form. Step 2 Integrating both sides, we get 𝑑𝑦 ∫ ln 𝑥 𝑑𝑥 = − ∫ 𝑦 ln 𝑦 1 Let ln 𝑦 = 𝑡 ⇒ 𝑑𝑦 = 𝑑𝑡 𝑦 Using integration by parts (ln 𝑥) ⋅ ∫ 1 𝑑𝑥 − ∫ ( 𝑑 (ln 𝑥) ⋅ ∫ 1𝑑𝑥) 𝑑𝑥 = − ln 𝑡 + 𝑐 𝑑𝑥 1 ⇒ (ln 𝑥)𝑥 − ∫ ⋅ 𝑥𝑑𝑥 = − ln 𝑡 + 𝑐 𝑥 ⇒ (ln 𝑥)𝑥 − ∫ 1 ⋅ 𝑑𝑥 = − ln 𝑡 + 𝑐 ⇒ 𝑥 ln 𝑥 − 𝑥 = − ln(ln 𝑦) + 𝑐 Question #17 (2𝑥𝑦 − 3𝑥 2 )𝑑𝑥 + (𝑥 2 + 𝑦)𝑑𝑦 = 0 Answer: The given differential equation is (2𝑥𝑦 − 3𝑥 2 )𝑑𝑥 + (𝑥 2 + 𝑦)𝑑𝑦 = 0 Comparing it with 𝑀𝑑𝑥 + 𝑀𝑑𝑦 = 0 𝑀 = 2𝑥𝑦 − 3𝑥 2 𝑁 = 𝑥 2 + 𝑦 𝑑𝑀 𝑑𝑁 = 2𝑥 = 2𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑀 Here, 𝑑𝑦 = 𝑑𝑁 𝑑𝑥 Then, it is exact differential equation, so solution is ∫ 𝑀 𝑑𝑥 + ∫ 𝑁 (term not contining x) 𝑑𝑦 = 𝑐 ∫(2𝑥 − 3𝑥 2 )𝑑𝑥 + ∫ 𝑦𝑑𝑦 = 𝑐 𝑦⋅ 2𝑥 2 3𝑥 3 𝑦 2 − + =𝑐 2 3 2 𝑥 2𝑦 − 𝑥 3 + 𝑦2 2 =𝑐 This is solution of given DE. Question #18 Topic: Exact Differential Equations Instructions: Solve the Exact Differential Equations of GIven Problem #2 only, ignore the other given problem. Note: Please answer all of the problems in the attached photo. I will rate you with “like/upvote” after; if you answer all of the problems. If not, I will rate you “unlike/downvote.” Kindly show the complete step-by-step solution. 1. (𝑥𝑒 𝑥 𝑦 + 𝑒 𝑥 𝑦)𝑑𝑥 + (8 + 𝑥𝑒 𝑥 )𝑑𝑦 = 0; 𝑦(0) = −4 2. ( 𝑥+𝑦 𝑦 2 +1 ) 𝑑𝑦 + (arctan 𝑦 + 𝑥)𝑑𝑥 = 0 Answer: (2) Given differential equation ( 𝑥+𝑦 𝑦 2 +1 ) 𝑑𝑦 + (arctan 𝑦 + 𝑥)𝑑𝑥 = 0 𝑥+𝑦 (tan1 (𝑦) + 𝑥)𝑑𝑥 + ( 2 ) 𝑑𝑦 = 0 𝑦 +1 Comparing with 𝑀𝑑𝑥 + 𝑁𝑑𝑦 = 0 𝑀 = tan−1 (𝑦) + 𝑥 𝜕𝑀 1 = +0 𝜕𝑦 1 + 𝑦 2 𝜕𝑀 1 = 𝜕𝑦 1 + 𝑦 2 𝑁= 𝑥+𝑦 𝑥 𝑦 = + 𝑦2 + 1 𝑦2 + 1 𝑦2 + 1 𝜕𝑁 1 = +0 𝜕𝑥 𝑦 2 + 1 𝜕𝑁 1 = 𝜕𝑥 1 + 𝑦 2 𝜕𝑀 𝜕𝑁 = 𝜕𝑦 𝜕𝑥 Given differential equation is exact differential equation. Silution of given differential equation ∫ 𝑛𝑑𝑥 + ∫ 𝑁(terms free from x)𝑑𝑦 = 𝑐 (1) 𝑦 1 2𝑦 ∫ 𝑁 (terms of N-free from x) = ∫ 𝑦2 +1 𝑑𝑦 = 2 ∫ 𝑦2 +1 1 from 1,2 and 3 = ln(𝑦 2 + 1) ⇒ (3) 2 [  f ( x) dx] f ( x) 2 x 1 = ln( f ( x)) + cx tan ( y ) + + ln( y 2 + 1) 2 2 1 2 = c 2 x − tan ( y ) + x + ln( y 2 + 1) 1 = 2c 2 x tan1 ( y ) + x 2 + ln( y 2 + 1) = c1 [ 2c = c1 ] Question #19 Solve the following differential equations. Use the method of Bernoulli’s Equation. 𝑥(2𝑥 3 + 𝑦)𝑑𝑦 − 6𝑦 2 𝑑𝑥 = 0 Answer: Step 1 General Bernoulli's equation can be of two type in which we have, 𝑑𝑥 𝑑𝑦 + 𝑝(𝑦)𝑥 = 𝑞(𝑦)𝑥 𝑛 where p(y) and q(y) are polynomials of y. Since this equation is non-linear, make suitable substitutions to make this equation in linear form and then solve the equation just like solving linear equation. After solving substitute back the variables. Step 2 Rearranging the equation we have, 6𝑦 2 𝑑𝑥 − 𝑥(2𝑥 3 + 𝑦)𝑑𝑦 = 0 𝑑𝑥 𝑥 𝑥4 − = 2 𝑑𝑦 6𝑦 3𝑦 Now rearranging more, −3 𝑑𝑥 𝑑𝑦 𝑥4 + 1 −1 = 2 3 2𝑦𝑥 𝑦 Substitute 𝑣 = 1 𝑥3 𝑑𝑥 𝑑𝑣 −3 𝑑𝑦 = 𝑑𝑦 𝑥4 So that equation becomes, 𝑑𝑣 1 −1 + 𝑣= 2 𝑑𝑦 2𝑦 𝑦 Now this is linear form of equation, so calculating its integrating factor, 𝐼. 𝐹. = 𝑒 1 ∫2𝑦𝑑𝑦 1 = 𝑒 2 log(𝑦) 1 = 𝑒 log(𝑦)2 = √𝑦 Step 3 Now multiply the integrating factor with equation and solve, √𝑦 𝑑𝑣 1 −1 + √𝑦. . 𝑣 = √𝑦. 2 𝑑𝑦 2𝑦 𝑦 −1 −3 𝑑𝑣 𝑦 2 + ⋅ 𝑣 = −𝑦 2 √𝑦 𝑑𝑦 2 It can also be written in the form of derivative as, −3 𝑑 (√𝑦𝑣) = −𝑦 2 𝑑𝑦 ∫ −3 𝑑 (√𝑦𝑣) = ∫ − 𝑦 2 𝑑𝑦 −1 √𝑦. 𝑣 = −2𝑦 2 + 𝐶 −1 𝑣 = 𝐶𝑦 2 + 2𝑦 −1 Now substitute back the value of v, 1 𝑥3 −1 = 𝐶𝑦 2 + 2𝑦 −1 Step 4 Now, calculating y as, 𝑦= 3𝐶√𝑥 9 (9𝐶𝑥 3 + 8) 𝑥 3 (9𝐶 2 𝑥 3 + 4) + 2 2 Question #20 Solve for the following problems using substitution suggested by the equation. (3𝑥 − 2𝑦 + 1)𝑑𝑥 + (3𝑥 − 2𝑦 + 3)𝑑𝑦 = 0 Answer: Step 1 We have the given differential equation as (3𝑥 − 2𝑦 + 1)𝑑𝑥 + (3𝑥 − 2𝑦 + 3)𝑑𝑦 = 0 On solving further, we get the result a (3𝑥 − 2𝑦 + 3)𝑑𝑦 = −(3𝑥 − 2𝑦 + 1)𝑑𝑥 𝑑𝑦 −(3𝑥 − 2𝑦 + 1) = (3𝑥 − 2𝑦 + 3) 𝑑𝑥 Now, let us consider that 3𝑥 − 2𝑦 = 𝑢 3−2 3− 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑢 = 𝑑𝑥 𝑑𝑥 𝑑𝑢 𝑑𝑦 =2 𝑑𝑥 𝑑𝑥 1 𝑑𝑢 2 𝑑𝑥 = (3 − ) Step 2 Therefore, 3− 𝑑𝑢 𝑑𝑥 =− 1 2 (3 − 𝑑𝑢 𝑑𝑥 )= −(𝑢+1) 𝑢+3 2𝑢+2 𝑢+3 𝑑𝑢 2𝑢 + 2 = 3+ 𝑑𝑥 𝑢+3 𝑑𝑢 3𝑢 + 9 + 2𝑢 + 2 = 𝑑𝑥 𝑢+3 𝑑𝑢 5𝑢 + 11 = 𝑑𝑥 𝑢+3 ( 𝑢+3 5𝑢+11 ) 𝑑𝑢 = 𝑑𝑥 On integrating both of the side, we get the result as ∫ 𝑢+3 𝑑𝑢 = ∫ 𝑑𝑥 5𝑢 + 11 1 5𝑢 + 11 + 4 ∫ 𝑑𝑢 = 𝑥 + 𝑐 5 5𝑢 + 11 1 1 (𝑢 + ln(5𝑢 + 11)) = 𝑥 + 𝑐 5 5 1 1 (𝑢 + ln(5𝑢 + 11)) = 𝑥 + 𝑐 5 5 𝑢 ln(5𝑢 + 11) + =𝑥+𝑐 5 25 On substituting back 𝑢 = 3𝑥 − 2𝑦 , we get the result as (3𝑥 − 2𝑦) ln(15𝑥 − 10𝑦) + =𝑥+𝑐 5 25 Hence, general solution of differential equation (3𝑥 − 2𝑦 + 1)𝑑𝑥 + (3𝑥 − 2𝑦 + 3)𝑑𝑦 = 0 is ln(15𝑥−10𝑦) 25 =𝑥+𝑐 . (3𝑥−2𝑦) 5 +