MCQ
સમીકરણ $x\frac{{dy}}{{dx}} + 3y = x$ નો ઉકેલ મેળવો.
- A${x^3}y + \frac{{{x^4}}}{4} + c = 0$
- ✓${x^3}y = \frac{{{x^4}}}{4} + c$
- C${x^3}y + \frac{{{x^4}}}{4} = 0$
- Dએકપણ નહી.
It is in the form of $\frac{{dy}}{{dx}} + Py = Q$
So, $I.F.$ $ = {e^{\int_{}^{} {Pdx} }} = {e^3}^{\int_{}^{} {\frac{1}{x}dx} } = {e^{3\log x}} ={x^3}$
Hence required solution is
$y{{x}^{3}}=\int_{{}}^{{}}{{{x}^{3}}1dx\frac{{{x}^{4}}}{4}}+c=>y{{x}^{3}}=\frac{{{x}^{4}}}{4}+c$
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