Differential Equations: An Introduction

# Differential Equations: An Introduction ## 1 Differential Equation An equation involving differentials or differential coefficients or derivatives is called a differential equation. For example, - $dy/dx = x + 3$ - $d^2y/ dx^2 + 3dy/dx + 2y = 0$ - $dz/dy + dz/dx (dy/dx)^2 + c =0$ - $y +...

# Differential Equations: An Introduction ## 1 Differential Equation An equation involving differentials or differential coefficients or derivatives is called a differential equation. For example, - $dy/dx = x + 3$ - $d^2y/ dx^2 + 3dy/dx + 2y = 0$ - $dz/dy + dz/dx (dy/dx)^2 + c =0$ - $y + xy(dz/dx) = k$ - $1 + (dy/dx)^3/2 = ρ(d^2y/dx^2)$ - $p = (d^2y/dx^2)/(dy/dx) + a$ - $dy/dx = x + (y/dx)^2/2$ These are all called ordinary differential equations. They can be defined as equations in which all the differential coefficients have reference to a single independent variable. There are also partial differential equations which are beyond our scope. ## 2 Order and Degree of Differential Equation The order of a differential equation is the order of the highest derivative which appears in it. For example, the order of the equation: $1 + (dy/dx)^3/2 = ρ (d^2y/dx^2)$ is two, as the highest derivative is $d^2y/dx^2$, which appears in it. Thus, this differential equation is called a differential equation of the second order. The degree of a differential equation is the degree of the highest derivative which appears in it after the equation has been made clear of the radicals and the fractions. From the examples, we see that (i) is of the first degree, (ii) is of the second degree, (iii) is of the first degree, (iv) is of the first order and second degree, (v) is of the second order and second degree, and (vi) is of the first order and second degree. ### 1.2.1 Solutions of Differential Equations A solution or integral of a differential equation is a relation that exists between the dependent and independent variables, and at the same time the derivatives obtained therefrom satisfy the given equation. For example, $y = sin x + c$ is a solution of the differential equation: $dy/dx = cos x$. ### 1.2.2 Kinds of Solutions - **General Solution or the Complete Primitive:** The solution of a differential equation which contains a number of arbitrary constants equal to the order of the differential equation is called the general solution or the complete primitive. For example, $y = c_1cos x + c_2 sin x$ is a general solution of the differential equation: $d^2y/dx^2 + y = 0$ We observe here that the differential equation under consideration is of the second order; therefore, its general solution shall contain two arbitrary constants. Such constants which appear in the general solution, depending upon the order of the equation, are called arbitrary constants. - **Particular Solution:** If particular values are given to the arbitrary constants in the general solution, then the solution so obtained is called the particular solution. For example, the complete primitive of (i) is: $y = A cos (x + B) = a cos x + b sin x$ If we assign the particular values 1, 0 to these constants *a* and *b*, then $y = cos x$ is defined as a Particular Integral. Another possible Particular Integral is $y = sin x$. - **Singular Solution:** The solution which cannot be derived from the complete primitive or general solution by assigning partial values to the arbitrary constants are called the Singular Solutions. For example, $y^2 = 4ax$ is a singular solution of the differential equation: $d^2y/dx^2 = x + a$. ### Examples 1. Find the differential equation of the family of the curves $y = Ae^{2x} + Be^{-2x}$ for different values of *A* and *B*. **Solution:** We have $y = Ae^{2x} + Be^{-2x}$ Differentiating, we get: $dy/dx = 2Ae^{2x} – 2Be^{-2x}$ And: $d^2 y/ dx^2 = 4Ae^{2x} + 4Be^{-2x} = 4(Ae^{2x} + Be^{-2x}) = 4y$ Required differential equation is: $d^2y/dx^2 = 4y$ 2. Find the differential equation of the system of curves $y = ax² + b cos nx + c$, where *a*, *b*, *c* are arbitrary constants. **Solution:** We have $y = ax² + bcos nx + c$ Differentiating: $dy/dx = 2ax - bnsin nx$ $d^2y/dx^2 = 2a-bn^2 cos nx$ $d^3y/dx^3 = bn^2 sin nx$ Eliminating *a* between (ii) and (iii), we get: $x^2(d^y/dx^2) – y (dy/dx) = bn(x cos nx – sin nx)$ Eliminating *b* between (iv) and (v) by dividing, we get: $d^3y/dx^3 – (d^2y/dx^2)(dy/dx)/(x^2y)=-x^2sin nx/(x cos nx – sin nx)$ This is the required differential equation. Note that here three constants have been eliminated and we have got a differential equation of third order. ## Exercises 1. Find the differential equation from $x² – y² + 2xy = 1$, where *λ* is a parameter. 2. Determine a differential equation from the equation $ax² + by² = 1$, where *a* and *b* are parameters. 3. From $x² +y² + 2ax + 2by + c = 0$, derive a differential equation not containing *a*, *b*, and *c*. 4. Form the differential equation of all circles of radius *r*. 5. Show that $v = A/r + B$ is a solution of $dv/dr^2 + (2/r)(dv/dr) = 0$. 6. Find the differential equation corresponding to $y = ae^{2x} + be^{-3x} + ce^x$ ## Answers 1. $(x^2- y^2-1)(x(dy/dx) + y) + 2xy (dy/dx-x) = 0$ ## 2 Differential Equations of First Order and First Degree ### 2.1 Equations of the First Order and First Degree The general form of an ordinary differential equation of the first order and first degree is: $M + N(dy/dx) = 0$ where *M* and *N* are functions of *x*, *y* i.e., *f(x, y)* and *(x, y)*, or are constants. This equation can also be expressed as: $dy/dx = f(x, y)/ φ(x, y)$ and: $φ(x, y)dy = f(x, y) dx$ and: $-f(x, y)dx + φ(x, y) dy = 0$ and: $M dx + Ndy = 0$ ### 2.2 Equations in Which the Variables are Separable If any differential equation can be written as: $f(x) dx + φ(y) dy = 0$ We say that variables are separable. Such equations can be solved as given below: $f(x)dx + φ(y)dy = 0$ $∫f(x) dx + ∫φ(y) dy = c$ where *c* is an arbitrary constant. **Working Rule:** 1. Arrange all the terms containing *x* together with *dx* on one side and all the terms of *y* with *dy* on the other, or put them inside the crooked brackets on one side only. 2. Integrate each term separately with respect to either *x* or *y* as associated with *dx* or *dy*. 3. Add an arbitrary constant on one side only. ### Examples 1. Solve: $y dy/dx-xy²/dx = (x + a)²$ **Solution:** The equation may be written as: $(y-ay²) dy/dx = (x+a)$ $y(1-ay) dy/dx = (x+a)$ Integrating, we get: $logy - log (1-ay) = log(x+a) + log c$ $log(y/(1-ay)) = logc(x + a)$ $ y = c(1 - ay)(x+a)$ 2. Solve: $3e^x tan y + (1 - e^x) sec^2y (dy/dx) = 0$ **Solution:** Put $tan y = v$ $sec^2y (dy/dv) = dv/dx$ The given equation becomes: $3e^xv + (1-e^x)(dv/dx) = 0$; or: $3e^x dx / (e^x - 1) = dv/v$ On integration: $3 log (e^x - 1) = log v + log c = log vc$ $(e^x - 1)^3 = cv$ or $(e^x - 1)^3 = c tan y$ ### Exercise Solve the following differential equations: 1. $dy/dx = (1+y^2)/(1+x²)$ 2. $dy/dx = -x + 2y$ 3. $dy/dx = -y^2 + x^2 + y$ 4. $dy/dx = x/(x+2)$ 5. a) $dy/dx = [x(2logx +1)]/[siny + y cos y]$ b) $dy/dx = [y (2log y + 1)]/[sinx + x cos x]$. 6. $(x²+x) dx + (x^2 + y) dy = 0$ ## Answers 1. $y-x = c(1+x)$ 2. $e^(y/2)=e^(x/2) + c$ 3. $ -e^(y/2) = e^(x/2) + c$ 4. $xy^2 = x^2 lox + c $ 5. (a) $ y sin y = x^2 log x + c$ (b) $x sin x = y^2 log y + c$ 6. $(x^2+1)(x^2+y^2) = c$ ### 2.3 Change of Variables Sometimes, a substitution or change of variables reduces a given differential equation to a form in which the variables are separable. The following examples will make the student familiar with the process: 1. Solve: $(x+y)² dy/dx = a²$ **Sol:** Put $x + y = v$, then $1 + dy/dx = dv/dx$ or $dy/dx = dv/dx -1$ Putting the values in the given equation, we have: $v² (dv/dx - 1) = a²$ $v²dv = (v² + a²) dx$ $v^2/(v^2 + a²) dv = dx$ $dx = v²/(v² + a² - a²) dv = v²/(v² + a²) dv = (1 - a²/(v² + a²))dv$ Integrating, we get: $x = v - a arctan (v/a) + c$ $x = x + y - a arctan((x+y)/a) + c$ 2. Solve: $x dy/dx – y dx/( x² + y²) = (x² + a²) dx$ **Sol.** Let $x = r cos θ$, $y = r sin θ$ $dx/dθ = dr/dθ cos θ - r sin θ$ $dy/dθ = dr/dθ sin θ + cos θ$ $dy/dx = (dr/dθ sin θ + rcos θ)/(dr/dθ cos θ – r sin θ)$ Also: $r (dr/dθ cos θ – r sin θ)/(rcos θ(dr/dθ sin θ + rcos θ)) = (dr/dθ cos θ – r sin θ)/(dr/dθ sin θ + rcos θ)$ $r^2 (cos² θ + sin² θ) = (dr/dθ)² = dr^2/dθ² = (a² - r²) / r²$ $dr^2 = (a² - r²) dθ²$ or $dr/((a² - r²)1/2 = dθ$ or $sin⁻¹ (r/a) = θ + c$ $r = a sin (θ + c)$ By putting the values of *r* and *θ,* we get: $(x^2 + y²)1/2 = a sin (tan⁻¹ (y/x) + c) $ ### Exercise Solve the following differential equations: 1. $(x-y)² dy/dx = a²$ 2. $dy/dx = sin(x+y) + cos(x+y)$ 3. $sin⁻¹ (y/x) dy/dx = x + y$ 4. $x dy/dx – y dx = (x² + y²) dx$ 5. $-x tan (y-x) = 1$ 6. $[(x+y – b) dy/(x+y + b)] + [(x+y + a)dx/(x+y- a)]$ = 0 7. $sin y dy/dx = cos y (1 - x cos y)$ 8. $sec²y + 2x tan y = x³, 2.4$ ### 2.4 Homogeneous Equations All the differential equations of the type: $dy/dx = [f₁(x, y)]/[f₂ (x, y)]$ where *f₁(x, y)* and *f₂(x, y)* are homogeneous functions of *x* and *y* of the same degree, are called homogenous differential equations. Such equations can be solved by putting $y = vx$, where *v* is a function of *x*. Then differentiating, we get: $dy/dx = v + x(dv/dx)$ Now the homogeneous equation takes up the new form: $v + x(dv/dx) = [ f₁(x, vx) ] / [f₂ (x, vx)]$ $v + x (dv/dx) = f(v)$ ($as the functions f₁(x, y) and f₂(x, y)$ are of the same degree, on simplification of $[ f₁(x, vx) ] / [f₂ (x, vx)]$, we would expect a function of *v* only [i.e., *f(v)*]. $dv/dx = [f(v) - v]/x$ Therefore: $∫dv/[f(v) - v] = ∫dx/x$ Integrating it, we have: $F(v) = logx + c$, if $c = log(1/a) = -logc$ $log(a/y) = F(x/y)$ **Working Rule:** 1. Put $y = vx$ (or $x = vy$). 2. Transform the given equation in terms of *v* and *x* (or *v* and *y*). 3. Apply the method of separation of variables. 4. In the end, after integration, replace *v* by $y/x$ (or $x/y$). ### Examples 1. Solve: $x²ydx – [x³ + y³]dy = 0$ **Sol:** The given equation is: $dy/dx = x²y/(x³ + y³)$ This is a homogeneous differential equation. To solve it, put $y = vx$: $dy/dx = v + x(dv/dx)$ $x^2 vx / (x^3 + v³x^3)= v + x(dv/dx)$ $v/ (1+v³) = v + x(dv/dx)$ $v(dv/dx) = -(1+v³) / x$ $dv/[v(1+v³)] = -dx/x$ Integrating, we get: $1/3v³ – log v = log x + log c$ $log cxy³ = 3v³$ or $log cx³ = 3y³/x³$ $cy³ = e^(x³/3y³) or cy = e^(x³/3y³)$ 2. Solve: $x^2dy/dx = y + x √(x² + y²)$. **Sol.:** The given equation is: $dy/dx = (y + x√(x² + y²))/ (x²)$ $dy/dx = (y/x) + √(1 + (y/x)²) $ Put: $y = vx$, $dv/dx = v + x(dv/dx)$: $v + x(dv/dx) = v+ √(1 + v²)$ $dv/(√(1+v²)) = dx$ Integrating, we get: $sinh⁻¹ (v) = x + c$ $v = sinh (x + c)$ $y = x sinh (x + c)$. ### Exercise Solve the following differential equations: 1. $(x² - y²)dx + 2xydy = 0$ 2. $2xy dx + (y^2 -x²) dy = 0$ 3. $(x² + y²) dx + 2xydy = 0$ 4. $(x^2+xy)dy = (x² + y²)dx$ 5. x(x-y)dy + y^2 dx= 0 6. $cxy = e^(-x/y)$ 7. $(x√(x² + y²) - y²)dx + xydy = 0$ 8. $y²(dy/dx) + x²(dy/dx) - xy (dy/dx) = 0$ 9. $x²(dy/dx) + y(x+y)dx=0$ 10. $x^2(dy/dx) = y(x +y)/2$ 11. $(2x - y) dx + (x-2y) dy = 0$ 12. $(1 + e^(x/y)) dx + e^(x/y)[(1-y/x)] dy = 0$ ## 2.5 Equations Reducible to a Homogeneous Form All the differential equations of the type: $dy/dx = (ax + by + c) / (Ax + By + C)$ can be reduced to the homogenous form as below: Put $X = x-h$ and $Y = y-k$ ⇒ $dX = dx$ and $dY = dy$ Therefore: $dy/dx = dY/dX$ $dy/dx = (ax + by + c)/(Ax + By + C)$ Then the equation $dy/dx = (ax+by +c)/(Ax + By +C)$ takes up the form: $dY/dX = (ax + by + (ah + bk + c))/(AX + BY + (Ah + Bk + C))$ Now, choose *h* and *k* so that: $ah + bk + c = 0$ $Ah + Bk + C = 0$ $dY/dX = (aX + bY)/ (AX + BY)$ This is a homogeneous differential equation. It can be solved as in the previous article. **Working Rule:** 1. Put $X = x + h$, $Y = y + k$ in the given differential equation. 2. Equate the constant terms in the numerator and denominator to zero. 3. Solve these resulting equations of *h* and *k* obtained in (2) for *h* and *k*. 4. Solve the resulting homogeneous equations in *X* and *Y* by the method of 2.4. 5. Replace *X* by *x - h* and *Y* by *y - k*. 6. Substitute the values of *h* and *k*. **(a) When a/b = A/B** (a) Put $ax + by = v$. (b) Solve the resulting equation by using 2.2 (variables are separable method). ### Examples 1. Solve: $(2x + y + 3) dx = (2y + x + 1) dy$: **Sol:** The given equation is: $dy/dx = (2x+y+3)/(2y+x+1)$ Here: $a = 2$, $b = 1$, $A = 1$, $B = 2$, $ab/AB = 1$ (case *A* above) This is the equation reducible to homogenous form. To solve it, put $x = x+h$ and $y = Y+k$: $dy/dX = (2X +Y + (2h + k + 3))/(2Y + X + (h + 2k + 1))$ Putting: $2h + k + 3 = 0$ $h + 2k + 1 = 0$ i.e: $h = -5/3$, $k =-1/3$. Hence, $x = X - 5/3$ and $y = Y - 1/3$ $dY/dX = (2X + Y)/(2Y + X)$ We get: $dY/dX = (2 + v)/(2v+1)$ $[(2v+1) dV/(2+v)] = dX$ $dV[(2v+1)/(2+v)(1-v)] = dX$ $dV[(-2v+ 1)/(1+v)] = 2dX/X $ Integrating, we get: $-3 log (1-v) - log (1+v) = 4 log X - log c$ $log X[(1+v)(1-v)³] = log c$ $(X+Y)(X - Y)³ = c$ Putting the values of $X$, $Y$ and *h*, *k* we get: $(x+y+1)(x-y+2)³ = c$ 2. Solve: $dy/dx = (x - y + 3)/(2x - 2y + 5)$. **Sol:** The given equation can be put as: $dy/dx = (x-y + 3)/(2 (x - y) + 5)$ Put: $x - y = v ⇒ dy = dx$ $dV/dx = 1 - (dy/dx)$ $dV = 1 - [(v+3)/(2v+5)]dx$ $dV = [2v + 5 - v - 3]/dx = (v + 2)/(2v + 5) dx$ $(2v+5)dV = (v+2) dx $ or: $(1 + 2/(v+2)) dv = dx$ On integrating, we get: $2v + log(v + 2) = x + c$ $2(x - y) + log(x-y+2) = x + c$ $x - 2y + log(x-y+2) = c$. ### Exercise Solve following differential equations: 1. $dy/dx = (2y - x - 4)/(y - 3x +3)$ 2. $dy/dx = (3y - x + 7)/(x - 7y - 3)$ 3. $(6x - 5y+4)dy + (y - 2x - 1) dx = 0$ 4. $(2x + 3y - 5)dx + (3x + 2y - 5)dy = 0$ 5. $dy/dx = (2x + 9y - 20)/(6x + 2y - 10)$ 6. $(y + x + 5) dy = (y - x + 1) dx$ 7. $dy/dx = (2x - y + 1)/(x + 2y - 3)$ 8. $(2x - y + 1) dx + (2y - x - 1) dy = 0$ 9. $dy/dx = (2x - 6y +7)/(x - 3y + 4)$ 10. $dy/dx = (2y + x - 1)/(2x + 4y + 3)$ 11. $dy/dx = (4x + 6y + 5)/(3y + 2x + 4)$ 12. $dy/dx = (x + y +1)/(2x + 2y + 3)$ ## 2.6 Linear Differential Equations The equation $dy/dx + Py = Q$, whose left side is linear in both the dependent variable and its derivative, is called a linear differential equation of the first order, when *P* and *Q* are the functions of *x* only or constants. The differential equation $dy/dx + Px = Q$, where *P*, *Q* are functions of *y* only or constants is also a linear differential equation (where *x* is the dependent variable). ### 2.6.1 Solution of Linear Equations The given equation is: $dy/dx + Py = Q$ $d/dx (ye^(∫Pdx)) = e^(∫Pdx) (dy/dx) + yPe^(∫Pdx)$ $d/dx (ye^(∫Pdx)) = e^(∫Pdx) (dy/dx + Py$ $ye^(∫Pdx) = ∫[Qe^(∫Pdx)] dx + c$ where $e^(∫Pdx)$ is an integrating factor of (i), and its primitive or general solution is: $ye^(∫Pdx) = ∫[Qe^(∫Pdx)] dx + c$ where *c* is an arbitrary constant. **Working Rule:** -(1) Determine the integrating factor $e^(∫Pdx)$ which is represented by I.F. -(2) Then multiply the differential equation by I.F. $e^(∫Pdx)$. Now integrate by using integration by parts method taking $e^(∫Pdx)$ as the first function and *y* as the second function. Then we get: **Dependent variable x I.F. = ∫[Qe^(∫Pdx)] dx + c** ** or y x I.F. = ∫[Qe^(∫Pdx)] dx + c** -(3) If *x* is the independent variable, then I.F. = $e^(∫Pdy)$ and the solution is: **x x I.F. = ∫[Qe^(∫Pdy)] dy + c** ### Examples 1. Solve: $(x+2)(dy/dx) = y$. **Sol:** The given equation can be written as: $dy/dx + (-1/(x+2))y = 0$ $dy/dx - y/(x+2) = 0$ It is of the form $dy/dx + Py = Q$ Hence, I.F. $= e^(∫-1/(x+2) dy) = e^(-logy) = 1/y $ (since *x* is the dependent variable). Therefore, the solution of the given equation will be: $(1/y) x = ∫[0/y]dy + c$ $x/y = c $ $x = cy$ 2. Solve: $(1+y²) dy/dx + (x - e^(tan⁻¹y)) y = 0$ **Sol:** The given equation is: $(1+y²) dy/dx + xy = e^(tan⁻¹y) (dy/dx)$ $(1 +y²) dy/dx + xy = e^(tan⁻¹y)$ $dy/dx + [x/(1+y²)]y = [e^(tan⁻¹y)]/(1+y²)$ I.F. $= e^(∫[x/(1+y²)]dx) = e^(tan⁻¹y) $ (since *x* is the dependent variable) Therefore, the solution will be: $xe^(tan⁻¹y) = ∫[e^(tan⁻¹y) * e^(tan⁻¹y)/(1+y²)] dy + c$ $xe^(tan⁻¹y) = ∫(e^(2tan⁻¹y)/(1+y²)) dy + c$ Let: $tan⁻¹y = t$, $1/(1+y²) dy = dt$ Then: $xe^(tan⁻¹y) = ∫e^(2t) dt + c$ $xe^(tan⁻¹y) = e^(2t)/2 + c$ $xe^(tan⁻¹y) = ½ e^(2tan⁻¹y) + c$ 3. Solve: $√(a²+x²) * dy/dx + y = √(a²+x²) - x$ **Sol:** The given equation can be put in the form: $dy/dx + (y/√(a²+x²)) = (√(a²+x²) - x)/√(a²+x²) $ $dy/dx + (y/√(a²+x²)) = 1 - (x/√(a²+x²))$ Hence, the solution of the given equation is: $y√(a² + x²) + x = ∫[√a²+x²) - (x/√(a²+x²))]√(a²+x²) dx + c$ $y√a²+x²) +x = ∫(a² + x² - x√(a²+x²)) = ∫(a²/(√(a²+x²))) dx + c$ $= a ∫(a²/(√(a²+x²))) dx = a sinh⁻¹ (x/a) + c$. ### Exercise Solve the following differential equations: 1. $dy/dx + (2x/(1+x²))y = 1/(1+x²)$ 2. $dy/dx + (y tan x) = sec x$ 3. $dy/dx + y tan x = sec x$ 4. $dy/dx + y tan x = - sec x$ 5. $dy/dx - y tan x = -2 sin x$ 6. $dy/dx + y cot x = 2 cos x$ 7. $dy/dx + y/x = x²$, if $y = 1$ when $x = 1$ 8. $(1+x²)dy/dx + y = tan⁻¹ x$ 9. $(1+y²) (x - e^(tan⁻¹y)) (dy/dx) = 0$ 10. $(1+ y²) dx = (tan⁻¹ y -x) dy$ 11. $x logx dy/dx + y = 2 log x$ 12. $(2x -10y³)dy/dx + y = 0$ 13. $xdy/dx - y = 2x² cosec 2x$ 14. $dy/dx + [y/(x+√(1-x²))] = (1+x²)/(1-x²)^(3/2)√(1-x²)$ 15. $cos³xdy/dx + y cos x = sin x$ ## 2.7 Differential Equations Reducible to the Linear Form (Bernoulli's Equation) A differential equation, having the form: $dy/dx + Py = Qy^n$ is called an equation reducible to the linear form (or Bernoulli’s equation), where *P* and *Q* are functions of *x* only. **Solution:** The given equation is: $dy/dx + Py = Qy^n$ $y^(-n)dy/dx + Py^(-n+1) = Q$ Putting $y^(-n+1) = v$: $(1-n)y^(-n) dy/dx = dv/dx$ $(-n + 1) y^(-n)dy/dx = dv/dx$ Then (i) becomes: $(1-n) dv/dx + Pv = Q$ $dv/dx - (n-1)Pv = Q(1-n)$ (ii) is a linear equation of the first order: Integrating factor is $e^(∫[-(n-1)P]dx)$ Therefore, the required solution is: $ve^(∫[-(n-1)P]dx) = ∫[Q(1-n)e^(∫[-(n-1)P]dx)] dx + c$ where *c* is an arbitrary constant. Similarly, $dy/dx + Px = Q₁x^n$, where *P* and *Q* are functions of *y* alone, can be reduced to the form as shown below: $ue^(∫[-(n-1)P]dy) = ∫[Q₁(1-n)e^(∫[-(n-1)P]dy)] dy + c$ where $u = x^(-n+1)$. **Working Rule:** -(A) 1. Write the equation in the form: $dy/dx + Py = Qy^n$ -(2) Divide by $y^n$ and bring it to the form as in § 2.7 equation (i) -(3) Put $ y^(-n+1) = v$ and change the equation to the form: $dv/dx – (n-1)Pv = Q(1-n)$ -(4) Then apply the § 2.7. -(B) In case of the equation: $dx/dy + Rx = Q₁x^n$ apply the similar line i.e., first divide by $x^n$, then put a substitution $x^(-n+1) = V$ and this brings a linear differential equatino (§ 2.6.1). ### Examples 1. Solve: $dy/dx – (tan y)/(1 + x) = (1+ x)e^x sec y$. **Sol:** The given equation is: $dy/dx – (tan y/1 + x) = (1 + x)e^x sec y$ $dy/dx – (sin y/ cos y (1 + x)) = (1 + x)e^x sec y$ $cos y(dy/dx) -siny / (1+ x) = (1 + x)e^x$ Let $sin y = v$, $cos y (dy/dx) = dv/dx$: $dv/dx – v/(1+x) = (1+x)e^x$ $I.F. = e^(∫-1/(1+x)dx) = e^(-log (1+x)) = 1/(1+x)$ Hence, the solution is: $1/(1+x) v = ∫[(1+x) * e^x] dx + c$ $1/(1+x)v = ∫[e^x + xe^x] dx + c$ $v/(1+x) = e^x + c$ $sin y = (1+x)(e^x + c)$