Find the solution of differential equation \(\left( {y + {x^2}} \right

1.	Find the solution of differential equation \(\left( {y + {x^2}} \right)\frac{{dx}}{{dy}} = x\)1. x = y2 + yc2. y = x + xc3. y = x2 + xc4. x = y + yc
Answer» Correct Answer - Option 3 : y = x² + xc 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 functions 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: \(\left( {y + {x^2}} \right)\frac{{dx}}{{dy}} = x\) \(x\frac{{dy}}{{dx}} = {x^2} + y\) \(\frac{{dy}}{{dx}} - \frac{y}{x} = x\) It is a form of \(\frac{{dy}}{{dx}} + Py = Q\) \(I.F. = {e^{\smallint pdx}}\) \(I.F. = {e^{ln\frac{1}{x}}} = \frac{1}{x}\) The solution of the linear equation is given by \(x\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c.\) \(y\left( {\frac{1}{x}} \right) = \smallint x\left( {\frac{1}{x}} \right)dx + c\) \(\frac{y}{x} = x + c\) \(y = {x^2} + xc\)

Find the solution of differential equation \(\left( {y + {x^2}} \right)\frac{{dx}}{{dy}} = x\)1. x = y2 + yc2. y = x + xc3. y = x2 + xc4. x = y + yc

Answer» Correct Answer - Option 3 : y = x² + xc

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 functions 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:

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

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

\(\frac{{dy}}{{dx}} - \frac{y}{x} = x\)

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

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

\(I.F. = {e^{ln\frac{1}{x}}} = \frac{1}{x}\)

The solution of the linear equation is given by

\(x\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c.\)

\(y\left( {\frac{1}{x}} \right) = \smallint x\left( {\frac{1}{x}} \right)dx + c\)