Question

Solve the following differential equations: (i) \(\frac{d y}{d x}=y \tan x-2 \sin x\) (ii) \(\left(1-x^{2}\right) \frac{d y}{d x}+2 x y=x\left(1-x^{2}\right)^{1 / 2}\) (iii) \((\mathrm{x}+\mathrm{a}) \frac{\mathrm{dy}}{\mathrm{dx}}-3 \mathrm{y}=(\mathrm{x}+\mathrm{a})^{5}\) (iv) \((x+1) \frac{d y}{d x}-n y=e^{x}(x+1)^{n+1}\).

Short answer

Separation of variables, Bernoulli's equation, integrating factors, and substitution are used to solve the given differential equations. 2. What is the integrating factor for the third differential equation? The integrating factor for the third differential equation is given by (x + a)^{-3}. 3. What substitution is used for the fourth differential equation? For the fourth differential equation, the substitution z = y(x + 1)^n is used.

Step-by-step solution

Separate the variables

We rewrite the equation as \(\frac{dy}{y}=(\tan x - 2\sin x) dx\).

Integrate both sides

Integrating left side with respect to \(y\) and right side with respect to \(x\), we have \(\int\frac{dy}{y} = \int(\tan x - 2\sin x) dx\).

Solve the integrals and find the solution

Integrating, we get \(\ln|y| = -\int 2\sin x dx + \int\tan x dx + C\). The integral of \(2\sin x\) is \(-2\cos x\), and the integral of \(\tan x\) is \(\ln|\sec x|\). Thus, \(\ln|y| = -2\cos x + \ln|\sec x| + C = \ln|\frac{y}{\sec x\ e^{2\cos x}}| = \ln|y\ e^{-2\cos x}\cos x| = C'\). Taking the exponential of both sides, we obtain the solution \(y = e^{C'}\sec x\ e^{2\cos x}\). (ii) \(\left(1-x^{2}\right) \frac{d y}{d x}+2 x y=x\left(1-x^{2}\right)^{\frac{1}{2}}\)

Identify it as a Bernoulli's equation

The equation can be rewritten as \(\frac{dy}{dx} + \frac{2x}{1-x^2}y = \frac{x\sqrt{1-x^2}}{1-x^2}\). This is a Bernoulli's equation with \(n=-1\).

Find the integrating factor

The integrating factor is given by \(e^{\int (2x)/(1-x^2) dx} = e^{\int \frac{2}{1+x} - \frac{2x^2}{1-x^2} dx} = e^{\ln|(1+x)| - x} = \frac{1+x}{e^x}\).

Multiply the equation by the integrating factor and integrate

Multiplying our equation by the integrating factor, we have \(\frac{1+x}{e^x}\frac{dy}{dx} + \frac{2x(1+x)}{e^x}y = \frac{x(1+x)\sqrt{1-x^2}}{e^x(1-x^2)}\). Notice the left side of the equation has the form \(\frac{d}{dx}\left(\frac{y(1+x)}{e^x}\right)\). So we can integrate both sides with respect to \(x\): \(\int\frac{d}{dx}\left(\frac{y(1+x)}{e^x}\right) dx = \int\frac{x(1+x)}{e^x}\sqrt{1-x^2} dx\).

Solve and find the solution

Integrating both sides, we have \(\frac{y(1+x)}{e^x} = \int\frac{x(1+x)}{e^x}\sqrt{1-x^2} dx + C\). To find the integral on the right side, we can use substitutions, numerical methods or look it up in a table of integrals. The solution will be in the form of: \(y(x) = e^x\left(\int\frac{x(1+x)}{e^x}\sqrt{1-x^2} dx + C\right) - xe^x\). (iii) \((\mathrm{x}+\mathrm{a}) \frac{\mathrm{dy}}{\mathrm{dx}}-3\mathrm{y}=(\mathrm{x}+\mathrm{a})^{5}\)

Identify it as a linear differential equation and find the integrating factor

The given equation is linear in y. The integrating factor is given by \(\mu(x)=e^{\int{-\frac{3}{x+a}}dx}=e^{-3\ln|x+a|}=(x+a)^{-3}\).

Multiply the equation by the integrating factor and integrate both sides

Multiplying the equation by \((x+a)^{-3}\), we get: \((x+a)^{-3}(x+a)\frac{dy}{dx}-3(x+a)^{-3}y=(x+a)^2\). Now the left side has the form \(\frac{d}{dx}\left((x+a)^{-3}y\right)\): \(\int\frac{d}{dx}\left((x+a)^{-3}y\right) dx = \int(x+a)^2 dx\).

Solve and find the solution

After integration, we have: \((x+a)^{-3}y = \frac{1}{4}(x^4+4x^3a+6x^2a^2+4a^3x+a^4)+C\). Multiplying both sides by \((x+a)^3\), we get the final solution: \(y(x) = \frac{1}{4}(x^4+4x^3a+6x^2a^2+4a^3x+a^4)(x+a)^3 + C(x+a)^3\). (iv) \((x+1) \frac{d y}{d x}-n y=e^{x}(x+1)^{n+1}\)

Identify a substitution and find the integrating factor

Let \(z = y(x+1)^n\). Then, \(\frac{dz}{dx} = (x+1)^n\frac{dy}{dx}+ny(x+1)^{n-1}\), and we can rewrite our equation as: \((x+1)\frac{dz}{dx}-z = e^{x}(x+1)^{n+1}\). The integrating factor for this equation is given by \(\mu(x)=e^{\int{-\frac{1}{x+1}}dx}=e^{-\ln|x+1|}=x+1\).

Multiply the equation by the integrating factor and integrate both sides

Multiplying the equation by the integrating factor, we get: \((x+1)^2\frac{dz}{dx}-(x+1)z=e^{x}(x+1)^{n+2}\). Now the left side has the form \(\frac{d}{dx}\left((x+1)^2z(x)\right)\): \(\int\frac{d}{dx}\left((x+1)^2z(x)\right) dx = \int e^{x}(x+1)^{n+2} dx\).

Solve and find the solution

After integration, we have: \((x+1)^2z(x) =\int e^{x}(x+1)^{n+2} dx + C\). Since \(z(x) = y(x+1)^n\), we can write the solution in terms of \(y(x)\): \(y(x) = (x+1)^{-n}\left(\int e^{x}(x+1)^{n+2} dx + C\right)\). To find the integral on the right side, we can use substitutions, numerical methods or look it up in a table of integrals.