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The general solution of the ode dydx+1xy=2x2𝑑𝑦𝑑𝑥+1𝑥𝑦=2𝑥2 isa.yx=x42+c𝑦𝑥=𝑥42+𝑐b.yx=x42𝑦𝑥=𝑥42c.yx=x33+c𝑦𝑥=𝑥33+𝑐d.yx=x44+c

Question

The general solution of the ode

dydx+1xy=2x2 \frac{dy}{dx} + \frac{1}{x}y = 2x^2

is

  • a. yx=x42+c yx = \frac{x^4}{2} + c
  • b. yx=x42y=x42 yx = \frac{x^4}{2}y = \frac{x^4}{2}
  • c. yx=x33+c yx = \frac{x^3}{3} + c
  • d. yx=x44+c yx = \frac{x^4}{4} + c
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Solution

The question seems to be a bit jumbled, but it appears to be asking for the general solution of a first order linear ordinary differential equation (ODE). The general form of such an equation is dy/dx + P(x)y = Q(x), where P(x) and Q(x) are some functions of x.

From the given equation, we can see that P(x) = 1/x and Q(x) = 2x^2.

The integrating factor e^(∫P(x)dx) = e^(∫1/x dx) = e^(ln|x|) = |x|.

Multiplying every term of the given differential equation by the integrating factor gives:

x dy/dx + y = 2x^3.

This is now in the form (d/dx)[x*y] = 2x^3, which can be integrated to give:

x*y = ∫2x^3 dx = 0.5x^4 + C.

So, the solution of the differential equation is y = 0.5x^3 + C/x, which is not exactly in the form of any of the provided options. There might be a mistake in the question or the provided options.

This problem has been solved

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