The solution of $$dydx+3xy=1x2$$ , given that $$y=2$$,$$x=1$$ is $$2x3y=x3+$$λ then λ is

I.Fe∫$$3dxx=e3log$$⁡$$x=x3$$ Solution of the differential equation is $$yx3=$$∫$$x3x2dx=x22+c$$ Pass (1.2)⇒$$c=3/2$$⇒$$yx3=x22+32$$⇒$$2x3y=x2+3$$⇒λ$$=3$$ .

Methods of solving first order first degree differential equations

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$The solution of \frac{d y}{d x} + \frac{3}{x} y = \frac{1}{x^{2}}, given that y = 2, x = 1 is 2 x^{3} y = x^{3} + λ then λ is$

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Solution

$\begin{array}{l} I . F e^{\int \frac{3 d x}{x}} = e^{3 \log x} = x^{3} \\ Solution of the differential equation is {yx}^{3} = \int \frac{x^{3}}{x^{2}} dx = \frac{x^{2}}{2} + c \\ Pass (1.2) \Rightarrow c = 3 / 2 \\ \Rightarrow {yx}^{3} = \frac{x^{2}}{2} + \frac{3}{2} \\ \Rightarrow 2 x^{3} y = x^{2} + 3 \\ \Rightarrow λ = 3 . \end{array}$

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Methods of solving first order first degree differential equations

Remember concepts with our Masterclasses.

The solution of dydx+3xy=1x2 , given that y=2,x=1 is 2x3y=x3+λ then λ is

I.Fe∫3dxx=e3log⁡x=x3 Solution of the differential equation is yx3=∫x3x2dx=x22+c Pass (1.2)⇒c=3/2⇒yx3=x22+32⇒2x3y=x2+3⇒λ=3 .

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$The solution of \frac{d y}{d x} + \frac{3}{x} y = \frac{1}{x^{2}}, given that y = 2, x = 1 is 2 x^{3} y = x^{3} + λ then λ is$