Question

Find the general solution of given differential equation \(\frac{{xdy}}{{dx}} + 3y = 4{x^3}\) ?
1. \({x^3}.y = \frac{2}{3}.{x^5} + c\)
2. \({x^3}.y = 3{x^6} + c\)
3. \({x^3}.y = \frac{2}{3}.{x^6} + c\)
4. None of the above
5. \({x^3}.y = \frac{2}{3}.{x^4} + c\)

Answer 1

Correct Answer - Option 3 : \({x^3}.y = \frac{2}{3}.{x^6} + c\)

Concept:

If a differential equation in the form of \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{Py}} = {\rm{Q}}\),

• Then we calculating integrating factor, I.F. = \({{\rm{e}}^{\smallint {\rm{P\;dx}}}}\)
• Solution will be \({\rm{y}}.{\rm{}}\left( {{\rm{I}}.{\rm{F}}.} \right){\rm{}} = {\rm{}}\smallint \left( {{\rm{Q}}.\left( {{\rm{I}}.{\rm{F}}.} \right)} \right){\rm{dx}} + {\rm{c}}\).

Calculation:

Given that,

⇒ \(\frac{{{\rm{xdy}}}}{{{\rm{dx}}}} + 3{\rm{y}} = {\rm{}}4{{\rm{x}}^3}\)

Now,

⇒ \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + \frac{{3{\rm{y}}}}{{\rm{x}}} = 4{{\rm{x}}^2}\)

By comparing with \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{Py}} = {\rm{Q}}\)

⇒ P = 3/x and Q = 4x²

⇒ I.F. = \({{\rm{e}}^{\smallint {\rm{P\;dx}}}} = {{\rm{e}}^{\smallint \frac{3}{{\rm{x}}}{\rm{\;dx}}}}\)

⇒ I.F. = \({{\rm{e}}^{3{\rm{lnx}}}}\)

⇒ I.F. = \({{\rm{e}}^{\ln {{\rm{x}}^3}}}\)

⇒ I.F. = x³ (∵ e^{ln x} = x)

Now general solution will be,

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

⇒ \({\rm{y}}.{\rm{}}\left( {{{\rm{x}}^3}} \right){\rm{}} = {\rm{}}\smallint \left( {4{{\rm{x}}^2}.\left( {{{\rm{x}}^3}} \right)} \right){\rm{dx}} + {\rm{c}}\)

⇒ \({{\rm{x}}^3}.{\rm{y}} = {\rm{}}\smallint 4{\rm{\;}}{{\rm{x}}^5}{\rm{dx}} + {\rm{c}}\)

⇒ \({{\rm{x}}^3}.{\rm{y}} = 4\frac{{{{\rm{x}}^6}}}{6} + {\rm{c}}\)

⇒ \({x^3}.y = \frac{2}{3}.{x^6} + c\)