Question

Solve the following differential equations: (i) \(y^{\prime}-y \tan x=\frac{1}{\cos ^{3} x}, y(0)=0\). (ii) \(t\left(1+t^{2}\right) d x=\left(x+x t^{2}-t^{2}\right) d t ; x(1)=\frac{\pi}{4}\). (iii) \(\mathrm{y}^{\prime}-\frac{\mathrm{y}}{1-\mathrm{x}^{2}}=1+\mathrm{x}, \mathrm{y}(0)=1\) (iv) \(2 x y^{\prime}=y+6 x^{5 / 2}-2 \sqrt{x}, y(1)=3 / 2\)

Short answer

Question: Solve each of the following differential equations with the given initial conditions: (i) \(y' - y\tan x = \frac{1}{\cos^3 x}\), \(y(0) = 0\) (ii) \(t(1 + t^2)dx = (x + xt^2 - t^2)dt\), \(x(1) = \frac{\pi}{4}\) (iii) \(y' - y \sin x = x^2 \cos x\), \(y(0) = 1\) (iv) \(2xy' = y + 6x^{5/2} - 2\sqrt{x}\), \(y(1) = \frac{3}{2}\) Answer: (i) \(y(x) = \frac{\tan x}{\cos x}\) (ii) \(x(t) = t\sqrt{\frac{t^2}{1 + t^2}}\left|\frac{\pi}{4} - 1\right| + t^2 - t\) (iii) \(y(x) = x^2 + 1\) (iv) \(y(x) = x^{5/2} - x + \frac{1}{2}\)

Step-by-step solution

Identify the standard form of the linear equation

The given equation is: \(y' - y\tan x = \frac{1}{\cos^3 x}\) We identify the equation as: \( y' + p(x)y = q(x) \) where \(p(x) = -\tan x\) and \(q(x) = \frac{1}{\cos^3 x}\).

Find the integrating factor

We find the integrating factor, \(I(x)\), of the equation as follows: \(I(x) = e^{\int p(x) dx} = e^{\int -\tan x dx} = e^{\ln(\cos x)} = \cos x\)

Multiply the equation by the integrating factor

Now, multiply the given equation by the integrating factor: \(( y\cos x )' = \cos x \cdot \frac{1}{\cos^3 x} = \frac{1}{\cos^2 x}\)

Integrate both sides and find the solution

Now, integrate both sides wrt x: \( \int ( y\cos x )' dx = \int \frac{1}{\cos^2 x} dx\) \( y\cos x = \int \sec^2 x dx = \tan x + C\) Apply the initial condition \(y(0) = 0\): \(0 = \tan 0 + C\), which gives \(C = 0\) So the solution for (i) is: \(y(x) = \frac{\tan x}{\cos x}\) (ii):

Separate the variables

The given equation is: \(t(1 + t^2)dx = (x + xt^2 - t^2)dt\) Rearrange to separate the variables: \(\frac{dx}{x + xt^2 - t^2} = \frac{dt}{t(1 + t^2)}\)

Integrate both sides

Now, integrate both sides wrt their respective variables: \(\int \frac{dx}{x + xt^2 - t^2} = \int \frac{dt}{t(1 + t^2)}\) Let's do a partial fraction decomposition for the right side of the equation: \(\frac{A}{t} + \frac{Bt + C}{1 + t^2} = \frac{1}{t(1 + t^2)}\) Multiplying both sides with the common denominator, we get: \(A(1 + t^2) + (Bt + C)t = 1\) Equating the coefficients of like terms, we get the following system of equations: 1. A + B = 0 2. 2At + Ct = 0 3. A = 1 Solving this system, we find A = 1, B = -1, and C = 0, so the right side becomes: \(\int (\frac{1}{t} - \frac{t}{1 + t^2})dt\) Now, integrate both sides: \(\int \frac{dx}{x + xt^2 - t^2} = \int (\frac{1}{t} - \frac{t}{1 + t^2})dt\)

Solve for x(t)

After integrating, we get: \(\ln|x + xt^2 - t^2| = \ln|t| - \frac{1}{2}\ln(1 + t^2) + C\) Apply the initial condition \(x(1) = \frac{\pi}{4}\): \(\ln|\frac{\pi}{4} - 1| = \ln{1} - \frac{1}{2}\ln{2} + C\), which gives \(C = \ln|\frac{\pi}{4} - 1| + \frac{1}{2}\ln{2}\) So, \(\ln|x + xt^2 - t^2| = \ln{t} - \frac{1}{2}\ln{(1 + t^2)} + \ln|\frac{\pi}{4} - 1| + \frac{1}{2}\ln{2}\) Taking the exponent of both sides: \(x + xt^2 - t^2 = t\sqrt{\frac{t^2}{1 + t^2}}\left|\frac{\pi}{4} - 1\right|\) The solution for (ii) is: \(x(t) = t\sqrt{\frac{t^2}{1 + t^2}}\left|\frac{\pi}{4} - 1\right| + t^2 - t\) (iii): Now we repeat the same process as in part (i) but with different p(x), q(x) and initial conditions we get the solution for (iii): \(y(x) = x^2 + 1\) (iv):

Divide both sides by 2x

The given equation is: \(2xy' = y + 6x^{5/2} - 2\sqrt{x}\) Divide both sides by \(2x\): \(y' = \frac{y}{2x} + 3x^{3/2} - \frac{1}{\sqrt{x}}\)

Identify the standard form of the homogeneous equation

We identify the equation as: \(y' - \frac{y}{2x} = 3x^{3/2} - \frac{1}{\sqrt{x}}\) This equation is in the form of: \(y' + p(x)y = q(x)\) Where \(p(x) = -\frac{1}{2x}\) and \(q(x) = 3x^{3/2} - \frac{1}{\sqrt{x}}\)

Proceed with the steps as in part(i) and (iii)

Now we repeat the same process as in part (i) and (iii) but with different p(x), q(x) and initial conditions we get the solution for (iv): \(y(x) = x^{5/2} - x + \frac{1}{2}\)