The solution of differential equation \(x\frac{{dy}}{{dx}} + y = {x^3}

1.	The solution of differential equation \(x\frac{{dy}}{{dx}} + y = {x^3}\) with the condition that y = 1 / 4 at x = 1 is1. \(y = {\frac{x}{3}^3}\)2. \(y = {\frac{x}{4}^3}\)3. \(x = {\frac{y}{3}^3}\)4. \(x = {\frac{y}{4}^3}\)
Answer» Correct Answer - Option 2 : \(y = {\frac{x}{4}^3}\) Concept: The standard form of a linear equation of the first order is given by \(\frac{{dy}}{{dx}} + Py = Q\) where P, Q are arbitrary function of x. The integrating factor of the linear equation is given by \(I.F. = {e^{\smallint pdx}}\) The solution of the linear equation is given by \(y\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c.\) Calculation: \(x\frac{{dy}}{{dx}} + y = {x^3}\) \(\frac{{dy}}{{dx}} + \frac{y}{x} = {x^2}\) It is form of \(\frac{{dy}}{{dx}} + Py = Q\) Here, P = 1/x and Q = x² \(I.F. = {e^{\smallint pdx}}\) \(I.F. = {e^{lnx}} = x\) The solution of the linear equation is given by \(y\left( x \right) = \smallint {x^2}\left( x \right)dx + c\) \(xy = {\frac{x}{4}^4} + c\) Put initial condition x = 1 and y = 1 / 4 in the above equation we get ⇒ c = 0 \(xy = {\frac{x}{4}^4}\) \(y = {\frac{x}{4}^3}\)

The solution of differential equation \(x\frac{{dy}}{{dx}} + y = {x^3}\) with the condition that y = 1 / 4 at x = 1 is1. \(y = {\frac{x}{3}^3}\)2. \(y = {\frac{x}{4}^3}\)3. \(x = {\frac{y}{3}^3}\)4. \(x = {\frac{y}{4}^3}\)

Answer» Correct Answer - Option 2 : \(y = {\frac{x}{4}^3}\)

Concept:

The standard form of a linear equation of the first order is given by \(\frac{{dy}}{{dx}} + Py = Q\) where P, Q are arbitrary function of x.

The integrating factor of the linear equation is given by \(I.F. = {e^{\smallint pdx}}\)

The solution of the linear equation is given by \(y\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c.\)

Calculation:

\(x\frac{{dy}}{{dx}} + y = {x^3}\)

\(\frac{{dy}}{{dx}} + \frac{y}{x} = {x^2}\)

It is form of \(\frac{{dy}}{{dx}} + Py = Q\)

Here, P = 1/x and Q = x²

\(I.F. = {e^{\smallint pdx}}\)

\(I.F. = {e^{lnx}} = x\)

The solution of the linear equation is given by

\(y\left( x \right) = \smallint {x^2}\left( x \right)dx + c\)

\(xy = {\frac{x}{4}^4} + c\)

Put initial condition x = 1 and y = 1 / 4 in the above equation we get ⇒ c = 0