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(B) The given differential equation is $x \frac{dy}{dx} + y = bx^4$,which can be rewritten as $\frac{dy}{dx} + \frac{1}{x} y = bx^3$.This is a linear

Vedclass · Accepted Answer

$10$

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(B) The given differential equation is $x \frac{dy}{dx} + y = bx^4$,which can be rewritten as $\frac{dy}{dx} + \frac{1}{x} y = bx^3$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x}$ and $Q(x) = bx^3$.The integrating factor $I.F. = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$,which gives $yx = \int bx^3 \cdot x dx + C = \int bx^4 dx + C$.Thus,$yx = \frac{bx^5}{5} + C$,or $f(x) = y = \frac{bx^4}{5} + \frac{C}{x}$.Since the curve passes through $(1, 2)$,we have $2 = \frac{b}{5} + C$,so $C = 2 - \frac{b}{5}$.Given $\int_{1}^{2} f(x) dx = \frac{62}{5}$,we have $\int_{1}^{2} (\frac{bx^4}{5} + \frac{C}{x}) dx = [\frac{bx^5}{25} + C \ln x]_{1}^{2} = \frac{62}{5}$.Substituting the limits: $(\frac{32b}{25} + C \ln 2) - (\frac{b}{25} + 0) = \frac{31b}{25} + C \ln 2 = \frac{62}{5}$.Substituting $C = 2 - \frac{b}{5}$: $\frac{31b}{25} + (2 - \frac{b}{5}) \ln 2 = \frac{62}{5}$.For this to hold,we set the coefficient of $\ln 2$ to $0$,so $2 - \frac{b}{5} = 0$,which gives $b = 10$.Then $\frac{31(10)}{25} = \frac{310}{25} = \frac{62}{5}$,which satisfies the equation. Thus,$b = 10$.

If a curve $y = f(x)$ passes through the point $(1, 2)$ and satisfies $x \frac{dy}{dx} + y = bx^4$,then for what value of $b$ is $\int_{1}^{2} f(x) dx = \frac{62}{5}$?

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