If ∫eaxcos⁡$$bx=e2x29f(x)+C$$ and f′′(x)=kf(x) then k is equal to

We have, ∫eaxcos⁡$$bxdx=eaxa2+b2(acos$$⁡$$bx+bsin$$⁡$$bx)+C$$⇒$$e2x22+52f(x)+C=eaxa2+b2(acos$$⁡$$bx+bsin$$⁡$$bx)+C$$⇒$$a=2$$,$$b=5$$ and $$f(x)=acos$$⁡$$bx+bsin$$⁡bx⇒f′′(x)=$$−$$b2f(x)=$$−$$25f(x)∴|k|$$=25$$

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If ${∫e}^{ax} \cos bx = \frac{e^{2 x}}{29} f (x) + C$ and $f^{''} (x) = kf (x)$ then $|k|$ is equal to

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$\begin{array}{l} We have, \int e^{ax} \cos bxdx = \frac{e^{ax}}{a^{2} + b^{2}} (acos bx + bsin bx) + C \\ \Rightarrow \frac{e^{2 x}}{2^{2} + 5^{2}} f (x) + C = \frac{e^{ax}}{a^{2} + b^{2}} (acos bx + bsin bx) + C \\ \Rightarrow a = 2, b = 5 and f (x) = acos bx + bsin bx \\ \Rightarrow f^{''} (x) = - b^{2} f (x) = - 25 f (x) \\ ∴ | k | = 25 \end{array}$

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Methods of integration

Remember concepts with our Masterclasses.

If ∫eaxcos⁡bx=e2x29f(x)+C and f′′(x)=kf(x) then k is equal to

We have, ∫eaxcos⁡bxdx=eaxa2+b2(acos⁡bx+bsin⁡bx)+C⇒e2x22+52f(x)+C=eaxa2+b2(acos⁡bx+bsin⁡bx)+C⇒a=2,b=5 and f(x)=acos⁡bx+bsin⁡bx⇒f′′(x)=−b2f(x)=−25f(x)∴|k|=25

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If ${∫e}^{ax} \cos bx = \frac{e^{2 x}}{29} f (x) + C$ and $f^{''} (x) = kf (x)$ then $|k|$ is equal to