If displaystyle ∫ e- (x2/2)dx=f(x) and the solution of the differential equation (d y/d x)=1+xy is y=ke(x2/2)f(x)+Ce(x2/2) , then the value of k is equal to (where C is the constant of integration)

Question

If displaystyle ∫ e- (x2/2)dx=f(x) and the solution of the differential equation (d y/d x)=1+xy is y=ke(x2/2)f(x)+Ce(x2/2) , then the value of k is equal to (where C is the constant of integration) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(d y/d x)-xy=1 Here, I.F. =e- displaystyle ∫ x d x=e- (x2/2) So, solution is ye- (x2/2)= displaystyle ∫ e- (x2/2)dx+C y=e(x2/2)f(x)+Ce(x2/2) Hence, k=1

Q. If $\int e^{- \frac{x ^{2}}{2}} d x = f (x)$ and the solution of the differential equation $\frac{d y}{d x} = 1 + x y$ is $y = k e^{\frac{x ^{2}}{2}} f (x) + C e^{\frac{x ^{2}}{2}}$ , then the value of $k$ is equal to (where $C$ is the constant of integration)

$\frac{d y}{d x} - x y = 1$
Here, I.F. $= e^{- \int x d x} = e^{- \frac{x ^{2}}{2}}$
So, solution is $y e^{- \frac{x ^{2}}{2}} = \int e^{- \frac{x ^{2}}{2}} d x + C$
$y = e^{\frac{x ^{2}}{2}} f (x) + C e^{\frac{x ^{2}}{2}}$
Hence, $k = 1$

Q. If ∫e−2x2​dx=f(x) and the solution of the differential equation dxdy​=1+xy is y=ke2x2​f(x)+Ce2x2​ , then the value of k is equal to (where C is the constant of integration)

dxdy​−xy=1 Here, I.F. =e−∫xdx=e−2x2​ So, solution is ye−2x2​=∫e−2x2​dx+C y=e2x2​f(x)+Ce2x2​ Hence, k=1

Q. If $\int e^{- \frac{x ^{2}}{2}} d x = f (x)$ and the solution of the differential equation $\frac{d y}{d x} = 1 + x y$ is $y = k e^{\frac{x ^{2}}{2}} f (x) + C e^{\frac{x ^{2}}{2}}$ , then the value of $k$ is equal to (where $C$ is the constant of integration)

$\frac{d y}{d x} - x y = 1$
Here, I.F. $= e^{- \int x d x} = e^{- \frac{x ^{2}}{2}}$
So, solution is $y e^{- \frac{x ^{2}}{2}} = \int e^{- \frac{x ^{2}}{2}} d x + C$
$y = e^{\frac{x ^{2}}{2}} f (x) + C e^{\frac{x ^{2}}{2}}$
Hence, $k = 1$