If the solution of the differential equationdy-dx-y-x-4-3x-2-x-2 -1-2-4x-3-e-x-3-x-2-1 is in the form y-f-x- -where f-0-1- then f-1-f-1- is -3-e- nbsp-2-e-4-e- nbsp-6-e- nbsp-

The correct option is D 6√(e) nbsp;I.F. = e^ ∫x^4 + 3x^2/(x^2 + 1)^2 = e^ x^3/x^2 + 1 y.e^ x^3/x^2 + 1 = ∫ (4x + 3)dx f(x) = y = (2x^2 + 3x + 1)e^x^3/x^2 ...

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Byju's Answer

Standard XII

Mathematics

Derivatives

If the soluti...

Question

If the solution of the differential equation $\frac{d y}{d x} - y (\frac{x^{4} + 3 x^{2}}{(x^{2} + 1)^{2}}) = (4 x + 3) e^{\frac{x^{3}}{x^{2} + 1}}$ is in the form $y = f (x)$ (where $f (0) = 1$ , then $f (1) + f (- 1)$ is -

2 \sqrt{e}

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3 \sqrt{e}

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4 \sqrt{e}

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6 \sqrt{e}

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Solution

The correct option is D $6 \sqrt{e}$
I.F. $= e^{- \int \frac{x^{4} + 3 x^{2}}{(x^{2} + 1)^{2}}} = e^{\frac{- x^{3}}{x^{2} + 1}} y . e^{\frac{- x^{3}}{x^{2} + 1}} = \int (4 x + 3) d x f (x) = y = (2 x^{2} + 3 x + 1) e^{\frac{x^{3}}{x^{2} + 1}} \Rightarrow f (1) = 6 \sqrt{e}$ & $f (- 1) = 0 \Rightarrow f (1) + f (- 1) = 6 \sqrt{e}$

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If the solution of the differential equationdydx−y(x4+3x2(x2+1)2)=(4x+3)ex3x2+1 is in the form y=f(x) (where f(0)=1, then f(1)+f(−1) is -

The correct option is D 6√e I.F. =e−∫x4+3x2(x2+1)2=e−x3x2+1y.e−x3x2+1=∫(4x+3)dxf(x)=y=(2x2+3x+1)ex3x2+1⇒f(1)=6√e & f(−1)=0⇒f(1)+f(−1)=6√e

If the solution of the differential equation $\frac{d y}{d x} - y (\frac{x^{4} + 3 x^{2}}{(x^{2} + 1)^{2}}) = (4 x + 3) e^{\frac{x^{3}}{x^{2} + 1}}$ is in the form $y = f (x)$ (where $f (0) = 1$ , then $f (1) + f (- 1)$ is -