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Differential Equations

Long Answer Type

331.

Solve the following differential equation:

$\cos^{2} x \frac{dy}{dx} + y = tanx$

332.

Solve the following differential equation:

$(1 + x^{2}) \frac{dy}{dx} + y = \tan^{- 1} x$

333.

Find the particular solution, satisfying the given condition, for the following differential equation:

$\frac{dy}{dx} - \frac{y}{x} + cosec (\frac{y}{x}) = 0; y = 0 when x = 1$

Short Answer Type

334.

What is the degree of the following differential equation?

5x $5 x {(\frac{dy}{dx})}^{2} - \frac{d^{2} y}{{dx}^{2}} - 6 y = \log x$

Long Answer Type

335.
Find the particular solution of the differential equation satisfying the given conditions: x² dy + (xy + y² )dx = 0; y = 1 when x = 1.

$x^{2} dy + (xy + y^{2}) dx = 0 x^{2} dy = - (xy + y^{2}) dx \Rightarrow \frac{dy}{dx} = - \frac{- (xy + y^{2})}{x^{2}} . . . . . . . . . . (i)$

This is a homogeneous differential equation.

Such type of equations can be reduced to variable separable form by the substitution y = vx.

Differentiating w.r.t.x we get,

$\frac{d}{dx} (y) = \frac{d}{dx} (vx) \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} substituting the value of y and \frac{dy}{dx} in equation (i), we get : v + x \frac{dv}{dx} = \frac{- [x \times vx + {(vx)}^{2}]}{x^{2}} \Rightarrow x \frac{dv}{dx} = - v^{2} - 2 v = - v (v + 2) \Rightarrow \frac{dv}{v (v + 2)} = - \frac{dx}{x} \Rightarrow \frac{1}{2} [\frac{1}{v} - \frac{1}{v + 2}] dv = - \frac{dx}{x}$

Integrating both sides, we get:

$\frac{1}{2} [\log v - \log (v + 2)] = - \log x + \log C \Rightarrow \frac{1}{2} \log (\frac{v}{v + 2}) = \log \frac{C}{x} \Rightarrow \frac{v}{v + 2} = {(\frac{C}{x})}^{2} Substituting v = \frac{y}{x}$

$\Rightarrow \frac{\frac{y}{x}}{\frac{y}{x} + 2} = {(\frac{C}{x})}^{2} \Rightarrow \frac{y}{y + 2 x} = \frac{C^{2}}{x^{2}} \Rightarrow \frac{x^{2} y}{y + 2 x} = D . . . . . . . . . . . . (ii)$

Now, it is given that y = 1 at x = 1.

$\Rightarrow \frac{1}{1 + 2} = D \Rightarrow D = \frac{1}{3} Substituting D = \frac{1}{3} in equation (ii), we get \frac{x^{2} y}{y + 2 x} = \frac{1}{3} \Rightarrow y + 2 x = 3 x^{2} y So, the required solution is y + 2 x = 3 x^{2} y$